Principles of Physics: A Calculus-Based Text

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FIFTH EDITION

Principles of Physics A CALCULUS-BASED TEXT Raymond A. Serway John W. Jewett, Jr.

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Principles of Physics, Fifth Edition Raymond A. Serway John W. Jewett, Jr. Publisher, Physical Sciences: Mary Finch Publisher, Physics and Astronomy: Charles Hartford Development Editor: Ed Dodd Associate Development Editor: Brandi Kirksey

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Printed in the United States of America 1 2 3 4 5 6 7 15 14 13 12 11

The MCAT Test Preparation Guide makes your copy of Principles of Physics, Fifth Edition, the most comprehensive MCAT study tool and classroom resource in introductory physics. The grid, which begins below and continues on the next two pages, outlines twelve concept-based study courses for the physics part of your MCAT exam. Use it to prepare for the MCAT, class tests, and your homework assignments.

Vectors

Force

Skill Objectives: To calculate distance, calculate angles between vectors, calculate magnitudes, and to understand vectors.

Skill Objectives: To know and understand Newton’s laws, to calculate resultant forces and weight. Review Plan:

Review Plan: Distance and Angles: Chapter 1 j Section 1.6 j Active Figure 1.4 j Chapter Problem 33 Using Vectors: Chapter 1 Sections 1.7–1.9 j Quick Quizzes 1.4–1.8 j Examples 1.6–1.8 j Active Figures 1.9, 1.16 j Chapter Problems 34, 35, 43, 44, 47, 51 j

Motion Skill Objectives: To understand motion in two dimensions, to calculate speed and velocity, to calculate centripetal acceleration, and acceleration in free fall problems. Review Plan: Motion in 1 Dimension: Chapter 2 j Sections 2.1, 2.2, 2.4, 2.6, 2.7 j Quick Quizzes 2.3–2.6 j Examples 2.1, 2.2, 2.4–2.9 j Active Figure 2.12 j Chapter Problems 3, 5,13, 19, 21, 29, 31, 33 Motion in 2 Dimensions: Chapter 3 j Sections 3.1–3.3 j Quick Quizzes 3.2, 3.3 j Examples 3.1–3.4 j Active Figures 3.5, 3.7, 3.10 j Chapter Problems 1, 11, 13 Centripetal Acceleration: Chapter 3 j Sections 3.4, 3.5 j Quick Quizzes 3.4, 3.5 j Example 3.5 j Active Figure 3.14 j Chapter Problems 23, 31

Newton’s Laws: Chapter 4 Sections 4.1–4.6 j Quick Quizzes 4.1–4.6 j Example 4.1 j Chapter Problem 7 j

Resultant Forces: Chapter 4 Section 4.7 j Quick Quiz 4.7 j Example 4.6 j Chapter Problems 29, 37 j

MCAT Test Preparation Guide

Welcome to your MCAT Test Preparation Guide

Gravity: Chapter 11 j Section 11.1 j Quick Quiz 11.1 j Chapter Problem 5

Equilibrium Skill Objectives: To calculate momentum and impulse, center of gravity, and torque. Review Plan: Momentum: Chapter 8 Section 8.1 j Quick Quiz 8.2 j Examples 8.2, 8.3 j

Impulse: Chapter 8 Sections 8.2–8.4 j Quick Quizzes 8.3, 8.4 j Examples 8.4, 8.6 j Active Figures 8.8, 8.9 j Chapter Problems 5, 9, 11, 17, 21 j

Torque: Chapter 10 Sections 10.5, 10.6 j Quick Quiz 10.7 j Example 10.8 j Chapter Problems 23, 30 j

iii

Work

Matter

Skill Objectives: To calculate friction, work, kinetic energy, power, and potential energy.

Skill Objectives: To calculate density, pressure, specific gravity, and flow rates.

Review Plan:

Review Plan:

Friction: Chapter 5 Section 5.1 j Quick Quizzes 5.1, 5.2 j

Work: Chapter 6 j Section 6.2 j Chapter Problems 3, 5 Kinetic Energy: Chapter 6 j Section 6.5 j Example 6.6 Power: Chapter 7 Section 7.6 j Chapter Problem 29

Density: Chapters 1, 15 j Sections 1.1, 15.2 Pressure: Chapter 15 j Sections 15.1–15.4 j Quick Quizzes 15.1–15.4 j Examples 15.1, 15.3 j Chapter Problems 2, 11, 23, 27, 31 Flow Rates: Chapter 15 j Section 15.6 j Quick Quiz 15.5

j

Potential Energy: Chapters 6, 7 Sections 6.6, 7.2 j Quick Quiz 6.6 Chapter 7 j Chapter Problem 3

MCAT Test Preparation Guide

j

iv

Waves Skill Objectives: To understand interference of waves, to calculate basic properties of waves, properties of springs, and properties of pendulums. Review Plan: Wave Properties: Chapters 12, 13 Sections 12.1, 12.2, 13.1, 13.2 j Quick Quiz 13.1 j Examples 12.1, 13.2 j Active Figures 12.1, 12.2, 12.6, 12.8, 12.11 Chapter 13 j Problem 7 j

Pendulum: Chapter 12 Sections 12.4, 12.5 j Quick Quizzes 12.5, 12.6 j Example 12.5 j Active Figure 12.13 j Chapter Problem 35 j

Interference: Chapter 14 j Sections 14.1, 14.2 j Quick Quiz 14.1 j Active Figures 14.1–14.3

Sound Skill Objectives: To understand interference of waves, calculate properties of waves, the speed of sound, Doppler shifts, and intensity. Review Plan: Sound Properties: Chapters 13, 14 j Sections 13.2, 13.3, 13.6, 13.7, 14.3 j Quick Quizzes 13.2, 13.4, 13.7 j Example 14.3 j Active Figures 13.6, 13.7, 13.9, 13.22, 13.23 Chapter 13 j Problems 11, 15, 28, 34, 41, 45 Chapter 14 j Problem 27 Interference/Beats: Chapter 14 j Sections 14.1, 14.5 j Quick Quiz 14.6 j Active Figures 14.1–14.3, 14.12 j Chapter Problems 9, 45, 46

Circuits

Skill Objectives: To understand mirrors and lenses, to calculate the angles of reflection, to use the index of refraction, and to find focal lengths.

Skill Objectives: To understand and calculate current, resistance, voltage, and power, and to use circuit analysis.

Review Plan:

Review Plan:

Reflection: Chapter 25 Sections 25.1–25.3 j Example 25.1 j Active Figure 25.5

Ohm’s Law: Chapter 21 Sections 21.1, 21.2 j Quick Quizzes 21.1, 21.2 j Examples 21.1, 21.2 j Chapter Problem 8

j

j

Refraction: Chapter 25 Sections 25.4, 25.5 j Quick Quizzes 25.2–25.5 j Example 25.2 j Chapter Problems 8, 16

j

j

Mirrors and Lenses: Chapter 26 j Sections 26.1–26.4 j Quick Quizzes 26.1–26.6 j Thinking Physics 26.2 j Examples 26.1–26.5 j Active Figures 26.2, 26.25 j Chapter Problems 27, 30, 33, 37

Power and Energy: Chapter 21 Section 21.5 j Quick Quiz 21.4 j Examples 21.4 j Active Figure 21.11 j Chapter Problems 21, 25, 31 Circuits: Chapter 21 Sections 21.6–21.8 j Quick Quizzes 21.5–21.7 j Examples 21.6–21.8 j Active Figures 21.14, 21.15, 21.17 j Chapter Problems 31, 39, 47

MCAT Test Preparation Guide

Light

j

Electrostatics Skill Objectives: To understand and calculate the electric field, the electrostatic force, and the electric potential.

Atoms Skill Objectives: To understand decay processes and nuclear reactions and to calculate half-life.

Review Plan: Coulomb’s Law: Chapter 19 j Sections 19.2–19.4 j Quick Quiz 19.1–19.3 j Examples 19.1, 19.2 j Active Figure 19.7 j Chapter Problems 3, 9 Electric Field: Chapter 19 Sections 19.5, 19.6 j Quick Quizzes 19.4, 19.5 j Active Figures 19.11, 19.20, 19.22 j

Potential: Chapter 20 Sections 20.1–20.3 j Examples 20.1, 20.2 j Active Figure 20.8 j Chapter Problems 3, 5, 8, 11 j

Review Plan: Atoms: Chapters, 11, 29 Section 11.5 j Sections 29.1–29.6 Chapter 11 j Problems 37–43, 61 j

Decays: Chapter 30 Sections 30.3, 30.4 j Quick Quizzes 30.3–30.6 j Examples 30.3–30.6 j Active Figures 30.8–30.11, 30.13, 30.14 j Chapter Problems 18, 23, 25 j

Nuclear Reactions: Chapter 30 Section 30.5 j Active Figure 30.18 j Chapter Problems 32, 35 j

v

Contents 4.6 4.7 4.8

About the Authors xi Preface xii To the Student xxviii Life Science Applications and Problems xxxi

An Invitation to Physics

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10

Introduction and Vectors

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

3 3.1 3.2 3.3 3.4 3.5 3.6 3.7

4 4.1 4.2 4.3 4.4 4.5 vi

5.1 5.2

4

Standards of Length, Mass, and Time 4 Dimensional Analysis 7 Conversion of Units 8 Order-of-Magnitude Calculations 9 Significant Figures 10 Coordinate Systems 12 Vectors and Scalars 13 Some Properties of Vectors 15 Components of a Vector and Unit Vectors 17 Modeling, Alternative Representations, and Problem-Solving Strategy 22

Context 1 | Alternative-Fuel Vehicles

2

5

1

5.3 5.4 5.5 5.6

6

35

Motion in One Dimension

37

Average Velocity 38 Instantaneous Velocity 41 Analysis Model: Particle Under Constant Velocity 45 Acceleration 47 Motion Diagrams 50 Analysis Model: Particle Under Constant Acceleration 51 Freely Falling Objects 56 Context Connection: Acceleration Required by Consumers 59

Motion in Two Dimensions

69

The Position, Velocity, and Acceleration Vectors 69 Two-Dimensional Motion with Constant Acceleration 71 Projectile Motion 74 Analysis Model: Particle in Uniform Circular Motion 81 Tangential and Radial Acceleration 83 Relative Velocity and Relative Acceleration 84 Context Connection: Lateral Acceleration of Automobiles 87

The Laws of Motion

97

The Concept of Force 97 Newton’s First Law 99 Mass 100 Newton’s Second Law 101 The Gravitational Force and Weight 104

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11

7 7.1 7.2 7.3 7.4 7.5 7.6 7.7

Newton’s Third Law 105 Analysis Models Using Newton’s Second Law 107 Context Connection: Forces on Automobiles 115

More Applications of Newton’s Laws 125 Forces of Friction 125 Extending the Particle in Uniform Circular Motion Model 130 Nonuniform Circular Motion 136 Motion in the Presence of Velocity-Dependent Resistive Forces 138 The Fundamental Forces of Nature 142 Context Connection: Drag Coefficients of Automobiles 144

Energy of a System

154

Systems and Environments 155 Work Done by a Constant Force 156 The Scalar Product of Two Vectors 158 Work Done by a Varying Force 160 Kinetic Energy and the Work–Kinetic Energy Theorem 165 Potential Energy of a System 168 Conservative and Nonconservative Forces 173 Relationship Between Conservative Forces and Potential Energy 175 Potential Energy for Gravitational and Electric Forces 176 Energy Diagrams and Equilibrium of a System 179 Context Connection: Potential Energy in Fuels 181

Conservation of Energy

192

Analysis Model: Nonisolated System (Energy) 193 Analysis Model: Isolated System (Energy) 195 Analysis Model: Nonisolated System in Steady State (Energy) 202 Situations Involving Kinetic Friction 203 Changes in Mechanical Energy for Nonconservative Forces 208 Power 214 Context Connection: Horsepower Ratings of Automobiles 216

Context 1 | CONCLUSION

Present and Future Possibilities 230 Context 2 | Mission to Mars

8 8.1 8.2 8.3

233

Momentum and Collisions

235

Linear Momentum 235 Analysis Model: Isolated System (Momentum) 237 Analysis Model: Nonisolated System (Momentum) 240

CONTENTS

vii

8.4 8.5 8.6 8.7 8.8

9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10

Collisions in One Dimension 243 Collisions in Two Dimensions 250 The Center of Mass 253 Motion of a System of Particles 257 Context Connection: Rocket Propulsion 260

Relativity

272

The Principle of Galilean Relativity 273 The Michelson–Morley Experiment 275 Einstein’s Principle of Relativity 276 Consequences of Special Relativity 276 The Lorentz Transformation Equations 285 Relativistic Momentum and the Relativistic Form of Newton’s Laws 288 Relativistic Energy 290 Mass and Energy 292 General Relativity 293 Context Connection: From Mars to the Stars 296

10 Rotational Motion

305

10.1 10.2

Angular Position, Speed, and Acceleration 306 Analysis Model: Rigid Object Under Constant Angular Acceleration 308 10.3 Relations Between Rotational and Translational Quantities 310 10.4 Rotational Kinetic Energy 311 10.5 Torque and the Vector Product 316 10.6 Analysis Model: Rigid Object in Equilibrium 320 10.7 Analysis Model: Rigid Object Under a Net Torque 323 10.8 Energy Considerations in Rotational Motion 326 10.9 Analysis Model: Nonisolated System (Angular Momentum) 328 10.10 Analysis Model: Isolated System (Angular Momentum) 331 10.11 Precessional Motion of Gyroscopes 335 10.12 Rolling Motion of Rigid Objects 336 10.13 Context Connection: Turning the Spacecraft 339

11 Gravity, Planetary Orbits, and the Hydrogen Atom 11.1 11.2 11.3 11.4 11.5 11.6

Context 3 | Earthquakes

388

12 Oscillatory Motion 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8

Motion of an Object Attached to a Spring 391 Analysis Model: Particle in Simple Harmonic Motion 392 Energy of the Simple Harmonic Oscillator 397 The Simple Pendulum 400 The Physical Pendulum 402 Damped Oscillations 403 Forced Oscillations 404 Context Connection: Resonance in Structures 405

13 Mechanical Waves 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8

Newton’s Law of Universal Gravitation Revisited 355 Structural Models 357 Kepler’s Laws 358 Energy Considerations in Planetary and Satellite Motion 364 Atomic Spectra and the Bohr Theory of Hydrogen 368 Context Connection: Changing from a Circular to an Elliptical Orbit 374

Context 2 | CONCLUSION

A Successful Mission Plan 384

415

Propagation of a Disturbance 416 Analysis Model: Traveling Wave 418 The Speed of Transverse Waves on Strings 423 Reflection and Transmission 426 Rate of Energy Transfer by Sinusoidal Waves on Strings 427 Sound Waves 429 The Doppler Effect 432 Context Connection: Seismic Waves 435

14 Superposition and Standing Waves 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8

447

Analysis Model: Waves in Interference 448 Standing Waves 451 Analysis Model: Waves Under Boundary Conditions 454 Standing Waves in Air Columns 456 Beats: Interference in Time 460 Nonsinusoidal Wave Patterns 462 The Ear and Theories of Pitch Perception 464 Context Connection: Building on Antinodes 466

Context 3 | CONCLUSION

Minimizing the Risk 476 Context 4 | Heart Attacks

15 Fluid Mechanics 354

390

15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9

479 482

Pressure 482 Variation of Pressure with Depth 484 Pressure Measurements 488 Buoyant Forces and Archimedes’s Principle 488 Fluid Dynamics 493 Streamlines and the Continuity Equation for Fluids 493 Bernoulli’s Equation 495 Other Applications of Fluid Dynamics 498 Context Connection: Turbulent Flow of Blood 499

Context 4 | CONCLUSION

Detecting Atherosclerosis and Preventing Heart Attacks 509

viii

CONTENTS Context 5 | Global Warming

513

16 Temperature and the Kinetic Theory of Gases 16.1 16.2 16.3 16.4 16.5 16.6 16.7

515

Temperature and the Zeroth Law of Thermodynamics 516 Thermometers and Temperature Scales 517 Thermal Expansion of Solids and Liquids 520 Macroscopic Description of an Ideal Gas 525 The Kinetic Theory of Gases 527 Distribution of Molecular Speeds 533 Context Connection: The Atmospheric Lapse Rate 535

17 Energy in Thermal Processes: The First Law of Thermodynamics 17.1 17.2 17.3 17.4 17.5 17.6 17.7 17.8 17.9 17.10 17.11

545

Heat and Internal Energy 546 Specific Heat 548 Latent Heat 550 Work in Thermodynamic Processes 554 The First Law of Thermodynamics 557 Some Applications of the First Law of Thermodynamics 559 Molar Specific Heats of Ideal Gases 562 Adiabatic Processes for an Ideal Gas 564 Molar Specific Heats and the Equipartition of Energy 566 Energy Transfer Mechanisms in Thermal Processes 568 Context Connection: Energy Balance for the Earth 573

18 Heat Engines, Entropy, and the Second Law of Thermodynamics 18.1 18.2 18.3 18.4 18.5 18.6 18.7 18.8 18.9

19.5 19.6 19.7 19.8 19.9 19.10 19.11 19.12

20 Electric Potential and Capacitance

Electric Potential and Potential Difference 657 Potential Difference in a Uniform Electric Field 658 Electric Potential and Potential Energy Due to Point Charges 661 20.4 Obtaining the Value of the Electric Field from the Electric Potential 664 20.5 Electric Potential Due to Continuous Charge Distributions 666 20.6 Electric Potential Due to a Charged Conductor 669 20.7 Capacitance 671 20.8 Combinations of Capacitors 674 20.9 Energy Stored in a Charged Capacitor 678 20.10 Capacitors with Dielectrics 681 20.11 Context Connection: The Atmosphere as a Capacitor 685

21 Current and Direct Current Circuits 21.1 21.2 21.3 21.4 21.5 21.6 21.7 21.8 21.9 21.10

Context 7 | Magnetism in Medicine

617

19 Electric Forces and 19.1 19.2 19.3 19.4

Electric Current 698 Resistance and Ohm’s Law 701 Superconductors 706 A Model for Electrical Conduction 707 Energy and Power in Electric Circuits 710 Sources of emf 713 Resistors in Series and Parallel 715 Kirchhoff ’s Rules 721 RC Circuits 724 Context Connection: The Atmosphere as a Conductor 729

Determining the Number of Lightning Strikes 739

Predicting the Earth’s Surface Temperature 612

Electric Fields

697

Context 6 | CONCLUSION

Context 5 | CONCLUSION

Context 6 | Lightning

656

20.1 20.2 20.3

586

Heat Engines and the Second Law of Thermodynamics 587 Reversible and Irreversible Processes 589 The Carnot Engine 589 Heat Pumps and Refrigerators 592 An Alternative Statement of the Second Law 593 Entropy 594 Entropy and the Second Law of Thermodynamics 597 Entropy Changes in Irreversible Processes 599 Context Connection: The Atmosphere as a Heat Engine 602

Electric Fields 627 Electric Field Lines 633 Motion of Charged Particles in a Uniform Electric Field 634 Electric Flux 636 Gauss’s Law 639 Application of Gauss’s Law to Various Charge Distributions 641 Conductors in Electrostatic Equilibrium 644 Context Connection: The Atmospheric Electric Field 645

22 Magnetic Forces and Magnetic Fields

619

Historical Overview 620 Properties of Electric Charges 620 Insulators and Conductors 622 Coulomb’s Law 624

741

22.1 22.2 22.3

743

Historical Overview 744 The Magnetic Field 745 Motion of a Charged Particle in a Uniform Magnetic Field 748

CONTENTS

ix

22.4

Applications Involving Charged Particles Moving in a Magnetic Field 751 22.5 Magnetic Force on a Current-Carrying Conductor 754 22.6 Torque on a Current Loop in a Uniform Magnetic Field 756 22.7 The Biot–Savart Law 758 22.8 The Magnetic Force Between Two Parallel Conductors 761 22.9 Ampère’s Law 762 22.10 The Magnetic Field of a Solenoid 766 22.11 Magnetism in Matter 767 22.12 Context Connection: Remote Magnetic Navigation for Cardiac Catheter Ablation Procedures 769

23 Faraday’s Law and Inductance 23.1 23.2 23.3 23.4 23.5 23.6 23.7 23.8

781

Faraday’s Law of Induction 781 Motional emf 786 Lenz’s Law 791 Induced emfs and Electric Fields 794 Inductance 796 RL Circuits 798 Energy Stored in a Magnetic Field 801 Context Connection: The Use of Transcranial Magnetic Stimulation in Depression 804

Context 7 | CONCLUSION

Nuclear Magnetic Resonance and Magnetic Resonance Imaging 817 Context 8 | Lasers

820

24 Electromagnetic Waves 24.1 24.2 24.3 24.4 24.5 24.6 24.7 24.8

822

Displacement Current and the Generalized Form of Ampère’s Law 823 Maxwell’s Equations and Hertz’s Discoveries 824 Electromagnetic Waves 826 Energy Carried by Electromagnetic Waves 830 Momentum and Radiation Pressure 833 The Spectrum of Electromagnetic Waves 836 Polarization of Light Waves 837 Context Connection: The Special Properties of Laser Light 839

25 Reflection and Refraction of Light 25.1 25.2 25.3 25.4 25.5 25.6 25.7 25.8

26 Image Formation by Mirrors and Lenses 26.1 26.2 26.3 26.4 26.5 26.6

879

Images Formed by Flat Mirrors 879 Images Formed by Spherical Mirrors 882 Images Formed by Refraction 888 Images Formed by Thin Lenses 891 The Eye 898 Context Connection: Some Medical Applications 900

27 Wave Optics 27.1 27.2 27.3 27.4 27.5 27.6 27.7 27.8 27.9 27.10

910

Conditions for Interference 911 Young’s Double-Slit Experiment 911 Analysis Model: Waves in Interference 913 Change of Phase Due to Reflection 916 Interference in Thin Films 916 Diffraction Patterns 919 Resolution of Single-Slit and Circular Apertures 922 The Diffraction Grating 925 Diffraction of X-Rays by Crystals 927 Context Connection: Holography 928

Context 8 | CONCLUSION

Using Lasers to Record and Read Digital Information 939 Context 9 | The Cosmic Connection

28 Quantum Physics

943

945

28.1 28.2 28.3 28.4 28.5 28.6 28.7 28.8 28.9 28.10 28.11

Blackbody Radiation and Planck’s Theory 946 The Photoelectric Effect 951 The Compton Effect 956 Photons and Electromagnetic Waves 959 The Wave Properties of Particles 959 A New Model: The Quantum Particle 963 The Double-Slit Experiment Revisited 965 The Uncertainty Principle 966 An Interpretation of Quantum Mechanics 968 A Particle in a Box 970 Analysis Model: Quantum Particle Under Boundary Conditions 974 28.12 The Schrödinger Equation 975 28.13 Tunneling Through a Potential Energy Barrier 977 28.14 Context Connection: The Cosmic Temperature 980

852

The Nature of Light 852 The Ray Model in Geometric Optics 853 Analysis Model: Wave Under Reflection 854 Analysis Model: Wave Under Refraction 857 Dispersion and Prisms 862 Huygens’s Principle 863 Total Internal Reflection 865 Context Connection: Optical Fibers 867

29 Atomic Physics 29.1 29.2 29.3 29.4 29.5 29.6 29.7

991

Early Structural Models of the Atom 992 The Hydrogen Atom Revisited 993 The Wave Functions for Hydrogen 996 Physical Interpretation of the Quantum Numbers 999 The Exclusion Principle and the Periodic Table 1004 More on Atomic Spectra: Visible and X-Ray 1008 Context Connection: Atoms in Space 1012

x

CONTENTS

30 Nuclear Physics 30.1 30.2 30.3 30.4 30.5 30.6

Some Properties of Nuclei 1022 Nuclear Binding Energy 1026 Radioactivity 1028 The Radioactive Decay Processes 1032 Nuclear Reactions 1039 Context Connection: The Engine of the Stars 1041

31 Particle Physics 31.1 31.2 31.3 31.4 31.5 31.6 31.7 31.8 31.9 31.10 31.11 31.12

1021

Context 9 | CONCLUSION

Problems and Perspectives 1086 Appendix A Tables A.1 A.1 Conversion Factors A.1 A.2 Symbols, Dimensions, and Units of Physical Quantities A.2 A.3 Chemical and Nuclear Information for Selected Isotopes A.4

Strange Particles and Strangeness 1065

Appendix B Mathematics Review A.6 B.1 Scientific Notation A.6 B.2 Algebra A.7 B.3 Geometry A.12 B.4 Trigonometry A.13 B.5 Series Expansions A.15 B.6 Differential Calculus A.15 B.7 Integral Calculus A.18 B.8 Propagation of Uncertainty A.22

Measuring Particle Lifetimes 1066

Appendix C Periodic Table of the Elements

1053

The Fundamental Forces in Nature 1054 Positrons and Other Antiparticles 1055 Mesons and the Beginning of Particle Physics 1057 Classification of Particles 1060 Conservation Laws 1061

Finding Patterns in the Particles 1067 Quarks 1069 Multicolored Quarks 1072 The Standard Model 1073 Context Connection: Investigating the Smallest System to Understand the Largest 1075

A.24

Appendix D SI Units A.26 D.1 SI Units A.26 D.2 Some Derived SI Units A.26 Answers to Quick Quizzes and Odd-Numbered Problems Index

I.1

A.27

About the Authors Raymond A. Serway

John W. Jewett, Jr.

xi

Preface Principles of Physics

xii

PREFACE

xiii What If?

| Objectives

| Changes in the Fifth Edition

New Contexts. 

Worked Examples.

Line-by-Line Revision of the Questions and Problems Set. 

Data from Enhanced WebAssign Used to Improve Questions and Problems. 

xiv

PREFACE

|A

Block Pulled on a Frictionless Surface n

v

F

SOLUTION Conceptualize

g

Figure 6.14  Categorize

Analyze

F

Finalize

Answer

What If?

PREFACE

xv

37.

35.

Figure P8.35 Figure P8.37

Revised Questions Organization. 

Objective Questions 

Conceptual Questions  Problems. 

black blue

red

xvi

PREFACE New Types of Problems.  Quantitative/Conceptual 

55.

Figure P7.55

Symbolic 

57.

Review.

v

Figure P7.57

57. Guided Problems 

PREFACE

xvii

28.

Figure P10.28

Impossibility Problems. 

51.

Increased Number of Paired Problems. 

Thorough Revision of Artwork. 

xviii

PREFACE

Henry Leap and Jim Lehman

Figure 10.28 

r

v

r

r

r

Figure 3.2  r

Expansion of the Analysis Model Approach. 

PREFACE

xix

Content Changes. 

| Organization

Context Number

Context

Physics Topics

Chapters

xx

PREFACE

| Text Features

Problem Solving and Conceptual Understanding General Problem-Solving Strategy. 

Thinking Physics. 

MCAT Test Preparation Guide. 

Active Figures. 

Quick Quizzes. 

QUICK QUIZ

a

Pitfall Prevention | 13.2

Two Kinds of Speed/Velocity

Pitfall Preventions. 

Summaries. 

b

c

d

e

PREFACE

xxi

Questions. 

Problems. 

Biomedical problems.

Paired Problems.

Review problems. Review

“Fermi problems.” Design problems. Calculus-based problems. www.cengage.com/physics/serway,

What If?

Alternative Representations. 

Math Appendix. 

Helpful Features Style. 

Important Definitions and Equations. 

boldface

xxii

PREFACE Marginal Notes.  Pedagogical Use of Color. 

c pedagogical color chart

Mathematical Level. 

Significant Figures. 

Units. 

Appendices and Endpapers. 

| TextChoice Custom Options for Principles of Physics

www. textchoice.com

www.cengage.com/custom

| Course Solutions That Fit Your Teaching Goals

and Your Students’ Learning Needs

PREFACE

xxiii

Homework Management Systems Enhanced WebAssign for

Fifth Edition. 

Master It tutorials

Master It tutorials

W

xxiv

PREFACE

Watch It

Cengage YouBook

www.webassign.net/ brookscole Quick Prep

PREFACE

xxv

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xxvi

PREFACE

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Student Solutions Manual/Study Guide 

Physics Laboratory Manual, Third Edition 

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| Teaching Options

PREFACE

xxvii

| Acknowledgments

Raymond A. Serway John W. Jewett, Jr.

To the Student It is appropriate

| How to Study

| Concepts and Principles

What If

| Study Schedule

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xxviii

TO THE STUDENT

xxix

| Use the Features boldface

| Problem Solving

xxx

TO THE STUDENT

| Experiments

| New Media Enhanced WebAssign

—Henri Poincaré

Life Science Applications and Problems Chapter 1: Introduction and Vectors 4

Context 4 Conclusion: Detecting Atherosclerosis and Preventing Heart Attacks 509–512

Chapter 2: Motion in One Dimension 37 Chapter 16: Temperature and the Kinetic Theory of Gases 515 Chapter 3: Motion in Two Dimensions 69 Chapter 4: The Laws of Motion 97 Chapter 5: More Applications of Newton’s Laws 125

Chapter 17: Energy in Thermal Processes: The First Law of Thermodynamics 545

Chapter 7: Conservation of Energy 192

Chapter 8: Momentum and Collisions 235 Chapter 18: Heat Engines, Entropy, and the Second Law of Thermodynamics 586 Chapter 9: Relativity 272 Chapter 19: Electric Forces and Electric Fields 619 Chapter 10: Rotational Motion 305

Chapter 20: Electric Potential and Capacitance 656

Context 2 Conclusion: A Successful Mission Plan 384 Chapter 12: Oscillatory Motion 390

Chapter 21: Current and Direct Current Circuits 697

Chapter 13: Mechanical Waves 415

Chapter 14: Superposition and Standing Waves 447

Context 7: Magnetism in Medicine 741–742 Chapter 22: Magnetic Forces and Magnetic Fields 743 Context 4: Heart Attacks 479–481 Chapter 15: Fluid Mechanics 482

Chapter 23: Faraday’s Law and Inductance 781

xxxi

xxxii

LIFE SCIENCE APPLICATIONS AND PROBLEMS

Context 7 Conclusion: Nuclear Magnetic Resonance and Magnetic Resonance Imaging 817–819 Chapter 27: Wave Optics 910 Chapter 24: Electromagnetic Waves 822

Chapter 28: Quantum Physics 945

Chapter 25: Reflection and Refraction of Light 852

Chapter 26: Image Formation by Mirrors and Lenses 879 Chapter 29: Atomic Physics 991

Chapter 30: Nuclear Physics 1021

Chapter 31: Particle Physics 1053

Stephen Inglis/Shutterstock.com

An Invitation to Physics

Stonehenge, in southern England, was built thousands of years ago. Various theories have been proposed about its function, including a burial ground, a healing site, and a place for ancestor worship. One of the more intriguing theories suggests that Stonehenge was an observatory, allowing for predictions of celestial events such as eclipses, solstices, and equinoxes.

P

hysics, the most fundamental physical science, is concerned with the basic principles of the universe. It is the foundation on which engineering, technology, and the other sciences — astronomy, biology, chemistry, and geology — are based. The beauty of physics lies in the simplicity of its fundamental theories and in the manner in which just a small number of basic concepts, equations, and assumptions can alter and expand our view of the world around us. Classical physics, developed prior to 1900, includes the theories, concepts, laws, and experiments in classical mechanics, thermodynamics, electromagnetism, and optics. For example, Galileo Galilei (1564 – 1642) made significant contributions to classical mechanics through his work on the laws of motion with constant acceleration. In the same era, Johannes Kepler (1571 – 1630) used astronomical observations to develop empirical laws for the motions of planetary bodies. The most important contributions to classical mechanics, however, were provided by Isaac Newton (1642 – 1727), who developed classical mechanics as a

systematic theory and was one of the originators of calculus as a mathematical tool. Although major developments in classical physics continued in the 18th century, thermodynamics and electromagnetism were not developed until the latter part of the 19th century, principally because the apparatus for controlled experiments was either too crude or unavailable until then. Although many electric and magnetic phenomena had been studied earlier, the work of James Clerk Maxwell (1831 – 1879) provided a unified theory of electromagnetism. In this text, we shall treat the various disciplines of classical physics in separate sections; we will see, however, that the disciplines of mechanics and electromagnetism are basic to all the branches of physics. A major revolution in physics, usually referred to as modern physics, began near the end of the 19th century. Modern physics developed mainly because many physical phenomena could not be explained by classical physics. The two most important developments in this modern era were the theories of relativity and quantum mechanics. Albert Einstein’s theory of relativity 1

© 2011 CERN

2  | An Invitation to Physics

The Compact Muon Solenoid (CMS) detector, part of the Large Hadron Collider operated by CERN (Conseil Européen pour la Recherche Nucléaire). The system is designed to detect and measure particles created in collisions of highenergy protons. Despite the word compact in the name, the detector is 15 meters in diameter. For a sense of scale, notice the worker in the blue helmet at the bottom of the photo as well as other workers in yellow helmets on the far side of the detector.

completely revolutionized the traditional concepts of space, time, and energy. This theory correctly describes the motion of objects moving at speeds comparable to the speed of light. The theory of relativity also shows that the speed of light is the upper limit of the speed of an object and that mass and energy are related. Quantum mechanics was formulated by a number of distinguished scientists to provide descriptions of physical phenomena at the atomic level. Scientists continually work at improving our understanding of fundamental laws, and new discoveries are made every day. In many research areas, a great deal of overlap exists among physics, chemistry, and biology. Evidence for this overlap is seen in the names of some subspecialties in science: biophysics, biochemistry, chemical physics, biotechnology, and so on. Numerous technological advances in recent times are the result of the efforts of many scientists, engineers, and technicians. Some of the most notable developments in the latter half of the 20th century were (1) space missions to the Moon and other planets, (2) microcircuitry and high-speed computers, (3) sophisticated imaging techniques used in scientific research and medicine, and (4) several remarkable accomplishments in genetic engineering. The early years of the 21st century have seen additional developments. Materials such as carbon nanotubes are now experiencing a variety of new applications. The 2010 Nobel Prize in Physics was

awarded for experiments performed on graphene, a two-dimensional material formed from carbon atoms. Potential applications include incorporation into a variety of electrical components and biodevices such as those used in DNA sequencing. The impact of such developments and discoveries on society has indeed been great, and future discoveries and developments will very likely be exciting, challenging, and of great benefit to humanity. To investigate the impact of physics on developments in our society, we will use a contextual approach to the study of the content in this textbook. The book is divided into nine Contexts, which relate the physics to social issues, natural phenomena, or medical/technological applications, as outlined here: Chapters

Context

2–7

Alternative-Fuel Vehicles

8–11

Mission to Mars

12–14

Earthquakes

15

Heart Attacks

16–18

Global Warming

19–21

Lightning

22–23

Magnetism in Medicine

24–27

Lasers

28–31

The Cosmic Connection

© 2011 Intuitive Surgical, Inc.

  | An Invitation to Physics 3

Physics is being used extensively today in the biomedical field. Shown here is the da Vinci Surgical System, a robotic device used to perform procedures such as prostatectomies, hysterectomies, mitral valve repairs, and coronary artery anastomosis. The surgeon sits at the console on the left and views a stereoscopic image of the surgery site. The movements of his hands are translated by a computer into movements of the robotic arms seen above the operating table at the right.

The Contexts provide a story line for each section of the text, which will help provide relevance and motivation for studying the material. Each Context begins with a discussion of the topic, culminating in a central question, which forms the focus for the study of the physics in the Context. The final section of each chapter is a Context Connection, in which the material in the chapter is explored with the central question in mind. At the end of each Context, a Context Conclusion brings together all the principles necessary to respond as fully as possible to the central question.

In Chapter 1, we investigate some of the mathematical fundamentals and problem-solving strategies that we will use in our study of physics. The first Context, AlternativeFuel Vehicles, is introduced just before Chapter 2; in this Context, the principles of classical mechanics are applied to the problem of designing, developing, producing, and marketing a vehicle that will help to reduce dependence on foreign oil and emit fewer harmful by-products into the atmosphere than current gasoline engines.

Chapter

1

Introduction and Vectors Chapter Outline 1.1

Standards of Length, Mass, and Time

1.2

Dimensional Analysis

1.3

Conversion of Units

1.4

Order-of-Magnitude Calculations

1.5

Significant Figures

1.6

Coordinate Systems

1.7

Vectors and Scalars

1.8

Some Properties of Vectors

1.9

Components of a Vector and Unit Vectors

1.10 Modeling, Alternative Representations, and Problem-Solving Strategy

T

he goal of physics is to provide a quantitative understanding of certain basic phenomena that occur in our A signpost in Saint Petersburg, Florida, Universe. Physics is a science based on shows the distance and direction to several experimental observations and mathematical analyses. The main objectives cities. Quantities that are defined by both a behind such experiments and analyses are to develop theories that explain the magnitude and a direction are called vector phenomenon being studied and to relate those theories to other established quantities. theories. Fortunately, it is possible to explain the behavior of various physical systems using relatively few fundamental laws. Analytical procedures require the expression of those laws in the language of mathematics, the tool that provides a bridge between theory and experiment. In this chapter, we shall discuss a few mathematical concepts and techniques that will be used throughout the text. In addition, we will outline an effective problem-solving strategy that should be adopted and Interactive content from this and other chapters may used in your problem-solving activities throughout the text. be assigned online in Enhanced WebAssign.

1.1 | Standards of Length, Mass, and Time To describe natural phenomena, we must make measurements associated with physical quantities, such as the length of an object. The laws of physics can be expressed as mathematical relationships among physical quantities that will be 4

Raymond A. Serway

SUMMARY

1.1 | Standards of Length, Mass, and Time 5

introduced and discussed throughout the book. In mechanics, the three fundamental quantities are length, mass, and time. All other quantities in mechanics can be expressed in terms of these three. If we measure a certain quantity and wish to describe it to someone, a unit for the quantity must be specified and defined. For example, it would be meaningless for a visitor from another planet to talk to us about a length of 8.0 “glitches” if we did not know the meaning of the unit glitch. On the other hand, if someone familiar with our system of measurement reports that a wall is 2.0 meters high and our unit of length is defined to be 1.0 meter, we then know that the height of the wall is twice our fundamental unit of length. An international committee has agreed on a system of definitions and standards to describe fundamental physical quantities. It is called the SI system (Système International) of units. Its units of length, mass, and time are the meter, kilogram, and second, respectively.

In a.d. 1120, King Henry I of England decreed that the standard of length in his country would be the yard and that the yard would be precisely equal to the distance from the tip of his nose to the end of his outstretched arm. Similarly, the original standard for the foot adopted by the French was the length of the royal foot of King Louis XIV. This standard prevailed until 1799, when the legal standard of length in France became the meter, defined as one ten-millionth of the distance from the equator to the North Pole. Many other systems have been developed in addition to those just discussed, but the advantages of the French system have caused it to prevail in most countries and in scientific circles everywhere. Until 1960, the length of the meter was defined as the distance between two lines on a specific bar of platinum – iridium alloy stored under controlled conditions. This standard was abandoned for several reasons, a principal one being that the limited accuracy with which the separation between the lines can be determined does not meet the current requirements of science and technology. The definition of the meter was modified to be equal to 1 650 763.73 wavelengths of orange – red light emitted from a krypton-86 lamp. In October 1983, the meter was redefined to be the distance traveled by light in a vacuum during a time interval of 1y299 792 458 second. This value arises from the establishment of the speed of light in a vacuum as exactly 299 792 458 meters per second. We will use the standard scientific notation for numbers with more than three digits in which groups of three digits are separated by spaces rather than commas. Therefore, 1 650 763.73 and 299 792 458 in this paragraph are the same as the more popular American cultural notations of 1,650,763.73 and 299,792,458. Similarly,  5 3.14159265 is written as 3.141 592 65.

© 2005 Geoffrey Wheeler Photography

Length

Figure 1.1 A cesium fountain atomic clock. The clock will neither gain nor lose a second in 20 million years.

c Definition of the meter

Mass Mass represents a measure of the resistance of an object to changes in its motion. The SI unit of mass, the kilogram, is defined as the mass of a specific platinum –iridium alloy cylinder kept at the International Bureau of Weights and Measures at Sèvres, France. At this point, we should add a word of caution. Many beginning students of physics tend to confuse the physical quantities called weight and mass. For the present we shall not discuss the distinction between them; they will be clearly defined in later chapters. For now you should note that they are distinctly different quantities.

Time Before 1967, the standard of time was defined in terms of the average length of a mean solar day. (A solar day is the time interval between successive appearances of the Sun at the highest point it reaches in the sky each day.) The basic unit of time, the second, was defined to be (1/60)(1/60)(1/24) 5 1/86 400 of a mean solar day. In 1967, the second was redefined to take advantage of the great precision obtainable with a device known as an atomic clock (Fig. 1.1), which uses the characteristic frequency of the

c Definition of the kilogram

6  CHAPTER 1 | Introduction and Vectors TABLE 1.2 |

TABLE 1.1 | Approximate Values of Some Measured Lengths Length (m)

Distance from the Earth to the most remote quasar

1.4 3 1026

Distance from the Earth to the most remote normal galaxies Distance from the Earth to the nearest large galaxy (M 31, the Andromeda galaxy)

9 3 1025 2 3 1022

Distance from the Sun to the nearest star (Proxima Centauri)

4 3 1016

Masses of Various Objects (Approximate Values) Mass (kg)

Visible Universe

, 1052 , 1042

One light-year

9.46 3 1015

Milky Way galaxy

Mean orbit radius of the Earth

1.50 3 1011

Sun

1.99 3 1030

Mean distance from the Earth to the Moon

3.84 3 108

Earth

5.98 3 1024

Distance from the equator to the North Pole

1.00 3

107

Moon

7.36 3 1022

Mean radius of the Earth

6.37 3 106

Shark

, 103

Human

, 102

Frog

, 1021

Mosquito

, 1025

Bacterium

, 10215

Typical altitude (above the surface) of a satellite orbiting the Earth

23

105

9.1 3 101

Length of a football field Length of this textbook

2.8 3

1021

5 3 1023

Length of a housefly

1024

Size of smallest visible dust particles

,

, 1025

Hydrogen atom

1.67 3 10227

Size of cells of most living organisms Diameter of a hydrogen atom

, 10210

Electron

9.11 3 10231

10214

, , 10215

Diameter of a uranium nucleus Diameter of a proton

c Definition of the second

Pitfall Prevention | 1.1

Reasonable Values Generating intuition about typical values of quantities when solving problems is important because you must think about your end result and determine if it seems reasonable. For example, if you are calculating the mass of a housefly and arrive at a value of 100 kg, this answer is unreasonable and there is an error somewhere.

c Definition of density

cesium-133 atom as the “reference clock.” The second is now defined as 9 192 631 770 times the period of oscillation of radiation from the cesium atom. It is possible today to purchase clocks and watches that receive radio signals from an atomic clock in Colorado, which the clock or watch uses to continuously reset itself to the correct time.

Approximate Values for Length, Mass, and Time Approximate values of various lengths, masses, and time intervals are presented in Tables 1.1, 1.2, and 1.3, respectively. Note the wide range of values for these quantities.1 You should study the tables and begin to generate an intuition for what is meant by a mass of 100 kilograms, for example, or by a time interval of 3.2 3 107 seconds. Systems of units commonly used in science, commerce, manufacturing, and everyday life are (1) the SI system, in which the units of length, mass, and time are the meter (m), kilogram (kg), and second (s), respectively; and (2) the U.S. customary system, in which the units of length, mass, and time are the foot (ft), slug, and second, respectively. Throughout most of this text we shall use SI units because they are almost universally accepted in science and industry. We will make limited use of U.S. customary units in the study of classical mechanics. Some of the most frequently used prefixes for the powers of ten and their abbreviations are listed in Table 1.4. For example, 1023 m is equivalent to 1 millimeter (mm), and 103 m is 1 kilometer (km). Likewise, 1 kg is 103 grams (g), and 1 megavolt (MV) is 106 volts (V). The variables length, time, and mass are examples of fundamental quantities. A much larger list of variables contains derived quantities, or quantities that can be expressed as a mathematical combination of fundamental quantities. Common examples are area, which is a product of two lengths, and speed, which is a ratio of a length to a time interval. Another example of a derived quantity is density. The density  (Greek letter rho; a table of the letters in the Greek alphabet is provided at the back of the book) of any substance is defined as its mass per unit volume: m ; 1.1b V 1 If

you are unfamiliar with the use of powers of ten (scientific notation), you should review Appendix B.1.

1.2 | Dimensional Analysis 7 TABLE 1.4 |

TABLE 1.3 | Approximate Values of Some Time Intervals

Some Prefixes for Powers of Ten

Time Interval (s)

4 3 1017

Age of the Universe Age of the Earth

Power

Prefix

Abbreviation

1.3 3 1017

10224

yocto

y

1012

10221

zepto

z

Average age of a college student

6.3 3 108

10218

atto

a

One year

3.2 3 107 8.6 3 104

10215

femto

f

10212

pico

p

1029

nano

n

1026

micro



8 3 1021

1023

milli

m

Period of audible sound waves

,

1023

1022

centi

c

Period of typical radio waves

, 1026

1021

deci

d

Period of vibration of an atom in a solid

, 10213

103

kilo

k

Period of visible light waves

, 10215

106

mega

M

Duration of a nuclear collision

, 10222

109

giga

G

Time interval for light to cross a proton

, 10224

1012

tera

T

1015

peta

P

1018

exa

E

1021

zetta

Z

1024

yotta

Y

Time interval since the fall of the Roman empire

One day (time interval for one revolution of the Earth about its axis) One class period

53

3.0 3

Time interval between normal heartbeats

103

which is a ratio of mass to a product of three lengths. For example, aluminum has a density of 2.70 3 103 kg/m3, and lead has a density of 11.3 3 103 kg/m3. An extreme difference in density can be imagined by thinking about holding a 10-centimeter (cm) cube of Styrofoam in one hand and a 10-cm cube of lead in the other.

1.2 | Dimensional Analysis In physics, the word dimension denotes the physical nature of a quantity. The distance between two points, for example, can be measured in feet, meters, or furlongs, which are all different ways of expressing the dimension of length. The symbols used in this book to specify the dimensions2 of length, mass, and time are L, M, and T, respectively. We shall often use square brackets [ ] to denote the dimensions of a physical quantity. For example, in this notation the dimensions of speed v are written [v] 5 L/T, and the dimensions of area A are [A] 5 L2. The dimensions of area, volume, speed, and acceleration are listed in Table 1.5, along with their units in the two common systems. The dimensions of other quantities, such as force and energy, will be described as they are introduced in the text. In many situations, you may be faced with having to derive or check a specific equation. Although you may have forgotten the details of the derivation, a useful and powerful procedure called dimensional analysis can be used as a consistency check, to assist in the derivation, or to check your final expression. Dimensional analysis makes use of the fact that dimensions can be treated as algebraic quantities. For example, quantities can be added or subtracted only if they have the same dimensions. Furthermore, the terms on both sides of an equation must have the same dimensions. By following these simple rules, you can use dimensional analysis to help determine TABLE 1.5 | Dimensions and Units of Four Derived Quantities Quantity

Area (A)

Volume (V )

Speed (v)

Acceleration (a)

Dimensions

L2

L3

L/T

L/T2

SI units

m2

m3

m/s

m/s2

U.S. customary units

ft2

ft3

ft/s

ft/s2

2 The dimensions of a variable will be symbolized by a capitalized, nonitalic letter, such as, in the case of length, L. The symbol for the variable itself will be italicized, such as L for the length of an object or t for time.

Pitfall Prevention | 1.2

Symbols for Quantities Some quantities have a small number of symbols that represent them. For example, the symbol for time is almost always t. Other quantities might have various symbols depending on the usage. Length may be described with symbols such as x, y, and z (for position); r (for radius); a, b, and c (for the legs of a right triangle); ℓ (for the length of an object); d (for a distance); h (for a height); and so forth.

8  CHAPTER 1 | Introduction and Vectors whether an expression has the correct form because the relationship can be correct only if the dimensions on the two sides of the equation are the same. To illustrate this procedure, suppose you wish to derive an expression for the position x of a car at a time t if the car starts from rest at t 5 0 and moves with constant acceleration a. In Chapter 2, we shall find that the correct expression for this special case is x 5 12at 2. Let us check the validity of this expression from a dimensional analysis approach. The quantity x on the left side has the dimension of length. For the equation to be dimensionally correct, the quantity on the right side must also have the dimension of length. We can perform a dimensional check by substituting the basic dimensions for acceleration, L/T2 (Table 1.5), and time, T, into the equation x 5 12at 2. That is, the dimensional form of the equation x 5 12at 2 can be written as L 2 T 5L T2 The dimensions of time cancel as shown, leaving the dimension of length, which is the correct dimension for the position x. Notice that the number 12 in the equation has no units, so it does not enter into the dimensional analysis. [x] 5

QUI C K QU IZ 1.1 True or False: Dimensional analysis can give you the numerical value of constants of proportionality that may appear in an algebraic expression.

Example 1.1 | Analysis

of an Equation

Show that the expression v 5 at, where v represents speed, a acceleration, and t an instant of time, is dimensionally correct.

SOLUTION Identify the dimensions of v from Table 1.5:

[v] 5

L T

Identify the dimensions of a from Table 1.5 and multiply by the dimensions of t:

[at] 5

L L T 5 T T2

Therefore, v 5 at is dimensionally correct because we have the same dimensions on both sides. (If the expression were given as v 5 at 2, it would be dimensionally incorrect. Try it and see!)

1.3 | Conversion of Units Pitfall Prevention | 1.3

Always Include Units When performing calculations, make it a habit to include the units with every quantity and carry the units through the entire calculation. Avoid the temptation to drop the units during the calculation steps and then apply the expected unit to the number that results for an answer. By including the units in every step, you can detect errors if the units for the answer are incorrect.

Sometimes it is necessary to convert units from one system to another or to convert within a system, for example, from kilometers to meters. Equalities between SI and U.S. customary units of length are as follows: 1 mile (mi) 5 1 609 m 5 1.609 km 1 m 5 39.37 in. 5 3.281 ft

1 ft 5 0.304 8 m 5 30.48 cm 1 inch (in.) 5 0.025 4 m 5 2.54 cm

A more complete list of equalities can be found in Appendix A. Units can be treated as algebraic quantities that can cancel each other. To perform a conversion, a quantity can be multiplied by a conversion factor, which is a fraction equal to 1, with numerator and denominator having different units, to provide the desired units in the final result. For example, suppose we wish to convert 15.0 in. to centimeters. Because 1 in. 5 2.54 cm, we multiply by a conversion factor that is the appropriate ratio of these equal quantities and find that 15.0 in. 5 (15.0 in.)

cm 5 38.1 cm 12.54 1 in. 2

1.4 | Order-of-Magnitude Calculations 9

where the ratio in parentheses is equal to 1. Notice that we express 1 as 2.54 cm/1 in. (rather than 1 in./2.54 cm) so that the inch cancels with the unit in the original quantity. The remaining unit is the centimeter, which is our desired result. QUICK QUIZ 1.2 The distance between two cities is 100 mi. What is the number of kilometers between the two cities? (a) smaller than 100 (b) larger than 100 (c) equal to 100

Exampl e 1.2 | Is

He Speeding?

On an interstate highway in a rural region of Wyoming, a car is traveling at a speed of 38.0 m/s. Is the driver exceeding the speed limit of 75.0 mi/h?

SOLUTION Convert meters in the speed to miles:

11 1609mim2 5 2.36 3 10

22

(38.0 m/s)

mi/s

5 85.0 mi/h 1160mins 2 1601min h 2

Convert seconds to hours: (2.36 3 1022 mi/s)

The driver is indeed exceeding the speed limit and should slow down.

What If? What if the driver were from outside the United States and is © Cengage Learning/Ed Dodd

familiar with speeds measured in kilometers per hour? What is the speed of the car in km/h? Answer We can convert our final answer to the appropriate units:

km 5 137 km/h 11.609 1 mi 2

(85.0 mi/h)

Figure 1.2 shows an automobile speedometer displaying speeds in both mi/h and km/h. Can you check the conversion we just performed using this photograph?

Figure 1.2 (Example 1.2) The speedometer of a vehicle that shows speeds in both miles per hour and kilometers per hour.

1.4 | Order-of-Magnitude Calculations Suppose someone asks you the number of bits of data on a typical musical compact disc. In response, it is not generally expected that you would provide the exact number but rather an estimate, which may be expressed in scientific notation. The estimate may be made even more approximate by expressing it as an order of magnitude, which is a power of ten determined as follows: 1. Express the number in scientific notation, with the multiplier of the power of ten between 1 and 10 and a unit. 2. If the multiplier is less than 3.162 (the square root of ten), the order of magnitude of the number is the power of ten in the scientific notation. If the multiplier is greater than 3.162, the order of magnitude is one larger than the power of ten in the scientific notation. We use the symbol , for “is on the order of.” Use the procedure above to verify the orders of magnitude for the following lengths: 0.008 6 m , 1022 m

0.002 1 m , 1023 m

720 m , 103 m

Usually, when an order-of-magnitude estimate is made, the results are reliable to within about a factor of ten. If a quantity increases in value by three orders of magnitude, its value increases by a factor of about 103 5 1 000.

10  CHAPTER 1 | Introduction and Vectors Example 1.3 | The

Number of Atoms in a Solid

Estimate the number of atoms in 1 cm3 of a solid.

SOLUTION From Table 1.1 we note that the diameter d of an atom is about 10210 m. Let us assume that the atoms in the solid are spheres of this diameter. Then the volume of each sphere is about 10230 m3 (more precisely, volume 5 4r 3/3 5 d 3/6, where r 5 d/2). Therefore, because 1 cm3 5 1026 m3, the number of atoms in the solid is on the order of 1026/10230 5 1024 atoms. A more precise calculation would require additional knowledge that we could find in tables. Our estimate, however, agrees with the more precise calculation to within a factor of 10.

Example 1.4 | Breaths

in a Lifetime

Estimate the number of breaths taken during an average human lifetime.

SOLUTION We start by guessing that the typical human lifetime is about 70 years. Think about the average number of breaths that a person takes in 1 min. This number varies depending on whether the person is exercising, sleeping, angry, serene, and so forth. To the nearest order of magnitude, we shall choose 10 breaths per minute as our estimate. (This estimate is certainly closer to the true average value than an estimate of 1 breath per minute or 100 breaths per minute.)

1

400 days

5 6 3 10 min 21125dayh 21601min h 2 5

Find the approximate number of minutes in a year:

1 yr

Find the approximate number of minutes in a 70-year lifetime:

number of minutes 5 (70 yr)(6 3 105 min/yr) 5 4 3 107 min

Find the approximate number of breaths in a lifetime:

number of breaths 5 (10 breaths/min)(4 3 107 min)

1 yr

5 4 3 108 breaths Therefore, a person takes on the order of 109 breaths in a lifetime. Notice how much simpler it is in the first calculation above to multiply 400 3 25 than it is to work with the more accurate 365 3 24.

What If? What if the average lifetime were estimated as 80 years instead of 70? Would that change our final estimate? Answer We could claim that (80 yr)(6 3 105 min/yr) 5 5 3 107 min, so our final estimate should be 5 3 108 breaths.

This answer is still on the order of 109 breaths, so an order-of-magnitude estimate would be unchanged.

1.5 | Significant Figures When certain quantities are measured, the measured values are known only to within the limits of the experimental uncertainty. The value of this uncertainty can depend on various factors, such as the quality of the apparatus, the skill of the experimenter, and the number of measurements performed. The number of significant figures in a measurement can be used to express something about the uncertainty. The number of significant figures is related to the number of numerical digits used to express the measurement, as we discuss below. As an example of significant figures, suppose we are asked to measure the radius of a compact disc using a meterstick as a measuring instrument. Let us assume the accuracy to which we can measure the radius of the disc is 60.1 cm. Because of the uncertainty of 60.1 cm, if the radius is measured to be 6.0 cm, we can claim only that its radius lies somewhere between 5.9 cm and 6.1 cm. In this case, we say that the

1.5 | Significant Figures 11

measured value of 6.0 cm has two significant figures. Note that the significant figures include the first estimated digit. Therefore, we could write the radius as (6.0 6 0.1) cm. Zeros may or may not be significant figures. Those used to position the decimal point in such numbers as 0.03 and 0.007 5 are not significant. Therefore, there are one and two significant figures, respectively, in these two values. When the zeros come after other digits, however, there is the possibility of misinterpretation. For example, suppose the mass of an object is given as 1 500 g. This value is ambiguous because we do not know whether the last two zeros are being used to locate the decimal point or whether they represent significant figures in the measurement. To remove this ambiguity, it is common to use scientific notation to indicate the number of significant figures. In this case, we would express the mass as 1.5 3 103 g if there are two significant figures in the measured value, 1.50 3 103 g if there are three significant figures, and 1.500 3 103 g if there are four. The same rule holds for numbers less than 1, so 2.3 3 1024 has two significant figures (and therefore could be written 0.000 23) and 2.30 3 1024 has three significant figures (also written as 0.000 230). In problem solving, we often combine quantities mathematically through multiplication, division, addition, subtraction, and so forth. When doing so, you must make sure that the result has the appropriate number of significant figures. A good rule of thumb to use in determining the number of significant figures that can be claimed in a multiplication or a division is as follows: When multiplying several quantities, the number of significant figures in the final answer is the same as the number of significant figures in the quantity having the smallest number of significant figures. The same rule applies to division. Let’s apply this rule to find the area of the compact disc whose radius we measured above. Using the equation for the area of a circle, A 5 r2 5 (6.0 cm)2 5 1.1 3 102 cm2 If you perform this calculation on your calculator, you will likely see 113.097 335 5. It should be clear that you don’t want to keep all of these digits, but you might be tempted to report the result as 113 cm2. This result is not justified because it has three significant figures, whereas the radius only has two. Therefore, we must report the result with only two significant figures as shown above. For addition and subtraction, you must consider the number of decimal places when you are determining how many significant figures to report: When numbers are added or subtracted, the number of decimal places in the result should equal the smallest number of decimal places of any term in the sum or difference. As an example of this rule, consider the sum 23.2 1 5.174 5 28.4 Notice that we do not report the answer as 28.374 because the lowest number of decimal places is one, for 23.2. Therefore, our answer must have only one decimal place. The rules for addition and subtraction can often result in answers that have a different number of significant figures than the quantities with which you start. For example, consider these operations that satisfy the rule: 1.000 1 1 0.000 3 5 1.000 4 1.002 2 0.998 5 0.004 In the first example, the result has five significant figures even though one of the terms, 0.000 3, has only one significant figure. Similarly, in the second calculation, the result has only one significant figure even though the numbers being subtracted have four and three, respectively.

Pitfall Prevention | 1.4

Read Carefully Notice that the rule for addition and subtraction is different from that for multiplication and division. For addition and subtraction, the important consideration is the number of decimal places, not the number of significant figures.

12  CHAPTER 1 | Introduction and Vectors c Significant figure guidelines

In this book, most of the numerical examples and end-of-chapter problems will yield answers having three significant figures. When carrying out estimation calculations, we shall typically work with a single significant figure.

used in this book

Pitfall Prevention | 1.5

Symbolic Solutions When solving problems, it is very useful to perform the solution completely in algebraic form and wait until the very end to enter numerical values into the final symbolic expression. This method will save many calculator keystrokes, especially if some quantities cancel so that you never have to enter their values into your calculator! In addition, you will only need to round once, on the final result.

Example 1.5 | Installing

If the number of significant figures in the result of a calculation must be reduced, there is a general rule for rounding numbers: the last digit retained is increased by 1 if the last digit dropped is greater than 5. (For example, 1.346 becomes 1.35.) If the last digit dropped is less than 5, the last digit retained remains as it is. (For example, 1.343 becomes 1.34.) If the last digit dropped is equal to 5, the remaining digit should be rounded to the nearest even number. (This rule helps avoid accumulation of errors in long arithmetic processes.) A technique for avoiding error accumulation is to delay the rounding of numbers in a long calculation until you have the final result. Wait until you are ready to copy the final answer from your calculator before rounding to the correct number of significant figures. In this book, we display numerical values rounded off to two or three significant figures. This occasionally makes some mathematical manipulations look odd or incorrect. For instance, looking ahead to Example 1.8 on page 21, you will see the operation 217.7 km 1 34.6 km 5 17.0 km. This looks like an incorrect subtraction, but that is only because we have rounded the numbers 17.7 km and 34.6 km for display. If all digits in these two intermediate numbers are retained and the rounding is only performed on the final number, the correct three-digit result of 17.0 km is obtained.

a Carpet

A carpet is to be installed in a rectangular room whose length is measured to be 12.71 m and whose width is measured to be 3.46 m. Find the area of the room.

SOLUTION If you multiply 12.71 m by 3.46 m on your calculator, you will see an answer of 43.976 6 m2. How many of these numbers should you claim? Our rule of thumb for multiplication tells us that you can claim only the number of significant figures in your answer as are present in the measured quantity having the lowest number of significant figures. In this example, the lowest number of significant figures is three in 3.46 m, so we should express our final answer as 44.0 m2.

1.6 | Coordinate Systems Many aspects of physics deal in some way or another with locations in space. For example, the mathematical description of the motion of an object requires a method for specifying the object’s position. Therefore, we first discuss how to describe the position of a point in space by means of coordinates in a graphical representation. A point on a line can be located with one coordinate, a point in a plane is located with two coordinates, and three coordinates are required to locate a point in space. A coordinate system used to specify locations in space consists of • A fixed reference point O, called the origin • A set of specified axes or directions with an appropriate scale and labels on the axes • Instructions that tell us how to label a point in space relative to the origin and axes

Figure 1.3 Designation of points in a Cartesian coordinate system. Each square in the xy plane is 1 m on a side. Every point is labeled with coordinates (x, y).

One convenient coordinate system that we will use frequently is the Cartesian coordinate system, sometimes called the rectangular coordinate system. Such a system in two dimensions is illustrated in Figure 1.3. An arbitrary point in this system is labeled with the coordinates (x, y). Positive x is taken to the right of the origin, and positive y is upward from the origin. Negative x is to the left of the origin, and negative y is downward from the origin. For example, the point P, which has coordinates (5, 3), may be reached by going first 5 m to the right of the origin and then 3 m above the origin

1.7 | Vectors and Scalars 13

(or by going 3 m above the origin and then 5 m to the right). Similarly, the point Q has coordinates (23, 4), which correspond to going 3 m to the left of the origin and 4 m above the origin. Sometimes it is more convenient to represent a point in a plane by its plane polar coordinates (r, ), as in Active Figure 1.4a. In this coordinate system, r is the length of the line from the origin to the point, and  is the angle between that line and a fixed axis, usually the positive x axis, with  measured counterclockwise. From the right triangle in Active Figure 1.4b, we find that sin  5 y/r and cos  5 x/r. (A review of trigonometric functions is given in Appendix B.4.) Therefore, starting with plane polar coordinates, one can obtain the Cartesian coordinates through the equations

u

x 5 r cos 

1.2b

y 5 r sin 

1.3b

u

Furthermore, if we know the Cartesian coordinates, the definitions of trigonometry tell us that

u u

y tan  5 x

1.4b

and r 5 "x 2 1 y 2

1.5b

You should note that these expressions relating the coordinates (x, y) to the coordinates (r, ) apply only when  is defined as in Active Figure 1.4a, where positive  is an angle measured counterclockwise from the positive x axis. Other choices are made in navigation and astronomy. If the reference axis for the polar angle  is chosen to be other than the positive x axis or if the sense of increasing  is chosen differently, the corresponding expressions relating the two sets of coordinates will change.

1.7 | Vectors and Scalars

Can you think of other examples? (b) This helpful person pointing in the correct direction tells us to travel five blocks north to reach the courthouse. A vector is a physical quantity that is specified by both magnitude and direction.

© Cengage Learning/George Semple

Mack Henley/Visuals Unlimited, Inc.

Each of the physical quantities that we shall encounter in this text can be placed in one of two categories, either a scalar or a vector. A scalar is a quantity that is completely specified by a positive or negative number with appropriate units. On the other hand, a vector is a physical quantity that must be specified by both magnitude and direction. The number of grapes in a bunch (Fig. 1.5a) is an example of a scalar quantity. If you are told that there are 38 grapes in the bunch, this statement completely specifies

Figure 1.5 (a) The number of grapes in this bunch is one example of a scalar quantity.

u

Active Figure 1.4 (a) The plane polar coordinates of a point are represented by the distance r and the angle , where  is measured in a counterclockwise direction from the positive x axis. (b) The right triangle used to relate (x, y) to (r, ).

14  CHAPTER 1 | Introduction and Vectors

c Displacement



훽 Figure 1.6 After a particle moves from 𝖠 to 𝖡 along an arbitrary path represented by the broken line, its displacement is a vector quantity shown by the arrow drawn from 𝖠 to 훾.

c Distance

Figure 1.7 A particle moving along the x axis from xi to xf undergoes a displacement Dx 5 xf 2 xi.

the information; no specification of direction is required. Other examples of scalars are temperature, volume, mass, and time intervals. The rules of ordinary arithmetic are used to manipulate scalar quantities; they can be freely added and subtracted (assuming that they have the same units!), multiplied and divided. Force is an example of a vector quantity. To describe the force on an object completely, we must specify both the direction of the applied force and the magnitude of the force. Another simple example of a vector quantity is the displacement of a particle, defined as its change in position. The person in Figure 1.5b is pointing out the direction of your desired displacement vector if you would like to reach a destination such as the courthouse. She will also tell you the magnitude of the displacement along with the direction, for example, “5 blocks north.” Suppose a particle moves from some point 𝖠 to a point 𝖡 along a straight path, as in Figure 1.6. This displacement can be represented by drawing an arrow from 𝖠 to 𝖡, where the arrowhead represents the direction of the displacement and the length of the arrow represents the magnitude of the displacement. If the particle travels along some other path from 𝖠 to 𝖡, such as the broken line in Figure 1.6, its displacement is still the vector from 𝖠 to 𝖡. The vector displacement along any indirect path from 𝖠 to 𝖡 is defined as being equivalent to the displacement represented by the direct path from 𝖠 to 𝖡. The magnitude of the displacement is the shortest distance between the end points. Therefore, the displacement of a particle is completely known if its initial and final coordinates are known. The path need not be specified. In other words, the displacement is independent of the path if the end points of the path are fixed. Note that the distance traveled by a particle is distinctly different from its displacement. The distance traveled (a scalar quantity) is the length of the path, which in general can be much greater than the magnitude of the displacement. In Figure 1.6, the length of the curved broken path is much larger than the magnitude of the solid black displacement vector. If the particle moves along the x axis from position xi to position xf , as in Figure 1.7, its displacement is given by xf 2 xi. (The indices i and f refer to the initial and final values.) We use the Greek letter delta (D) to denote the change in a quantity. Therefore, we define the change in the position of the particle (the displacement) as Dx ; xf 2 xi

1.6b

From this definition we see that Dx is positive if xf is greater than xi and negative if xf is less than xi. For example, if a particle changes its position from xi 5 2 5 m to xf 5 3 m, its displacement is Dx 5 1 8 m. Many physical quantities in addition to displacement are vectors. They include velocity, acceleration, force, and momentum, all of which will be defined in later chapters. In this text, we will use boldface letters with an arrow over the letter, such as : A , to represent vectors. Another common notation for vectors with which you should be familiar is a simple boldface character: A. : The magnitude of the vector A is written with an italic letter A or, alternatively, : u A u. The magnitude of a vector is always positive and carries the units of the quantity that the vector represents, such as meters for displacement or meters per second for velocity. Vectors combine according to special rules, which will be discussed in Sections 1.8 and 1.9.

QUICK QUIZ 1.3 Which of the following are vector quantities and which are scalar quantities? (a) your age (b) acceleration (c) velocity (d) speed (e) mass

1.8 | Some Properties of Vectors 15

THINKING PHYSICS 1.1

Consider your commute to work or school in the morning. Which is larger, the distance you travel or the magnitude of the displacement vector? Reasoning Unless you have a very unusual commute, the distance traveled must

be larger than the magnitude of the displacement vector. The distance includes the results of all the twists and turns you make in following the roads from home to work or school. On the other hand, the magnitude of the displacement vector is the length of a straight line from your home to work or school. This length is often described informally as “the distance as the crow flies.” The only way that the distance could be the same as the magnitude of the displacement vector is if your commute is a perfect straight line, which is highly unlikely! The distance could never be less than the magnitude of the displacement vector because the shortest distance between two points is a straight line. b

1.8 | Some Properties of Vectors Equality of Two Vectors :

:

Two vectors A and B are defined to be equal if they have the same units, the same : : : : magnitude, and the same direction. That is, A 5 B only if A 5 B and A and B point in the same direction. For example, all the vectors in Figure 1.8 are equal even though they have different starting points. This property allows us to translate a vector parallel to itself in a diagram without affecting the vector.

Figure 1.8 These four representations of vectors are equal because all four vectors have the same magnitude and point in the same direction.

Addition The rules for vector sums are conveniently described using a graphical method. : : : To add vector B to vector A , first draw a diagram of vector A on graph paper, with its : magnitude represented by a convenient scale, and then draw vector B to the same scale : with its tail starting from the tip of A , as in Active Figure 1.9a. The resultant vector : : : : : R 5 A 1 B is the vector drawn from the tail of A to the tip of B . The technique for adding two vectors is often called the “head-to-tail method.” When vectors are added, the sum is independent of the order of the addition. This independence can be seen for two vectors from the geometric construction in Active Figure 1.9b and is known as the commutative law of addition: :

:

:

:

1.7b

A1B5B1A

S

S

B

A

S

A

A

S

B

S

Pitfall Prevention | 1.6

S

S

B

A

S

R

S

B

S

B

S

R

S

A

S

B S

B

S

A

S

A

S

B

S

A

:

:

:

Active Figure 1.9 (a) When vector B is added to vector A, the resultant R is the :

:

vector that runs from the tail of A to the tip of B . (b) This construction shows that : : : : A 1 B 5 B 1 A ; vector addition is commutative.

Vector Addition Versus Scalar Addition :

:

:

Keep in mind that A 1 B 5 C is very different from A 1 B 5 C. The first equation is a vector sum, which must be handled carefully, such as with the graphical method described in Active Figure 1.9. The second equation is a simple algebraic addition of numbers that is handled with the normal rules of arithmetic.

16  CHAPTER 1 | Introduction and Vectors S

S

B

C S

A

:

S

S

C S

B

S

S

C

S

S

B S

A

S

S

A

S

B

C S

B

S

C

S

C

A

S

S

A

:

:

:

:

:

1.8b

A 1 (B 1 C) 5 (A 1 B) 1 C

C B

A

If three or more vectors are added, their sum is independent of the way in which they are grouped. A geometric demonstration of this property for three vectors is given in Figure 1.10. This property is called the associative law of addition:

S

B

S

Geometric constructions can also be used to add more than three vectors, : : as shown in Figure 1.11 for the case of four vectors. The resultant vector R 5 A 1 : : : B 1 C 1 D is the vector that closes the polygon formed by the vectors being added. : In other words, R is the vector drawn from the tail of the first vector to the tip of the last vector. Again, the order of the summation is unimportant. In summary, a vector quantity has both magnitude and direction and also obeys the laws of vector addition as described in Active Figure 1.9 and Figures 1.10 and 1.11. When two or more vectors are added together, they must all have the same units and they must all be the same type of quantity. It would be meaningless to add a velocity vector (for example, 60 km/h to the east) to a displacement vector (for example, 200 km to the north) because these vectors represent different physical quantities. The same rule also applies to scalars. For example, it would be meaningless to add time intervals to temperatures.

Negative of a Vector

B S

:

A

Figure 1.10 Geometric constructions for verifying the associative law of addition.

:

The negative of the vector A is defined as the vector that, when added to A , gives : : : : zero for the vector sum. That is, A 1 (2A) 5 0. The vectors A and 2A have the same magnitude but point in opposite directions.

Subtraction of Vectors S

The operation of vector subtraction makes use of the definition of the negative of a : : : : vector. We define the operation A 2 B as vector 2B added to vector A :

D

S

D

C

S

:

:

:

1.9b

The geometric construction for subtracting two vectors is illustrated in Figure 1.12.

R

S

A

S

B

S

C

:

A 2 B 5 A 1 (2B)

S

Multiplication of a Vector by a Scalar

S

B

:

S

A

Figure 1.11 Geometric construction for summing four vectors. The : resultant vector R closes the polygon and points from the tail of the first vector to the tip of the final vector.

Multiplication of Two Vectors :

S

S

A

:

Two vectors A and B can be multiplied in two different ways to produce either a sca: : lar or a vector quantity. The scalar product (or dot product) A ? B is a scalar quantity : : equal to AB cos , where  is the angle between A and B . The vector product (or cross : : product) A 3 B is a vector quantity whose magnitude is equal to AB sin . We shall discuss these products more fully in Chapters 6 and 10, where they are first used.

S

B

:

If a vector A is multiplied by a positive scalar quantity s, the product sA is a vector : that has the same direction as A and magnitude sA. If s is a negative scalar quantity, the : : : vector s A is directed opposite to A . For example, the vector 5A is five times longer : : 1: than A and has the same direction as A ; the vector 2 3 A has one-third the magni: : tude of A and points in the direction opposite A .

B

S

A

S

S

A

:

B

S

B

S

S

B

A

S

B

S

A

:

:

Figure 1.12 Subtracting vector B :

:

from vector A . The vector 2B is : equal in magnitude to vector B and points in the opposite direction.

:

QUICK QUIZ 1.4 The magnitudes of two vectors A and B are A 5 12 units and B 5 8 units. Which pair of numbers represents the largest and smallest possible values : : : for the magnitude of the resultant vector R 5 A 1 B ? (a) 14.4 units, 4 units (b) 12 units, 8 units (c) 20 units, 4 units (d) none of these answers :

QUICK QUIZ 1.5 If vector B is added to vector A , under what condition does the : : : : resultant vector A 1 B have magnitude A 1 B ? (a) A and B are parallel and in the : : : : same direction. (b) A and B are parallel and in opposite directions. (c) A and B are perpendicular.

1.9 | Components of a Vector and Unit Vectors 17

1.9 | Components of a Vector and Unit Vectors The graphical method of adding vectors is not recommended whenever high accuracy is required or in three-dimensional problems. In this section, we describe a method of adding vectors that makes use of the projections of vectors along coordinate axes. These projections are called the components of the vector or its rectangular components. Any vector can be completely described by its components. : Consider a vector A lying in the xy plane and making an arbitrary angle  with the positive x axis as shown in Figure 1.13a. This vector can be expressed as : : the sum of two other component vectors Ax , which is parallel to the x axis, and Ay, which is parallel to the y axis. From Figure 1.13b, we see that the three vectors form a right : : : triangle and that A 5 Ax 1 Ay. We shall often refer to the “components of a vector : A ,” written Ax and Ay (without the boldface notation). The component Ax rep: resents the projection of A along the x axis, and the component Ay represents : the projection of A along the y axis. These components can be positive or negative. : The component Ax is positive if the component vector Ax points in the positive x : direction and is negative if Ax points in the negative x direction. A similar statement is made for the component Ay. From Figure 1.13b and the definition of the sine and cosine of an angle, we see : that cos  5 Ax/A and sin  5 Ay/A. Hence, the components of A are given by Ax 5 A cos 

Ay 5 A sin 

and

Pitfall Prevention | 1.7

x and y Components Equation 1.10 associates the cosine of the angle with the x component and the sine of the angle with the y component. This association is true only because we measured the angle  with respect to the x axis, so do not memorize these equations. If  is measured with respect to the y axis (as in some problems), these equations will be incorrect. Think about which side of the triangle containing the components is adjacent to the angle and which side is opposite and then assign the cosine and sine accordingly.

1.10b

The magnitudes of these components are the lengths of the two sides of a right tri: angle with a hypotenuse of length A. Therefore, the magnitude and direction of A are related to its components through the expressions A 5 "Ax2 1 Ay2 tan  5

Ay Ax

1.11b

c Magnitude of

1.12b

c Direction of

:

:

To solve for , we can write  5 tan21 (Ay/Ax), which is read “ equals the angle whose tangent is the ratio Ay/Ax.” Note that the signs of the components Ax and Ay depend on the angle . For example, if  5 1208, Ax is negative and Ay is positive. If  5 2258, both Ax and Ay are negative. Figure 1.14 summarizes the signs of the components when : A lies in the various quadrants. If you choose reference axes or an angle other than those shown in Figure 1.13, the components of the vector must be modified accordingly. In many applications, it is more convenient to express the components of a vector in a coordinate system having axes that are not horizontal and vertical but are still perpendicular to each other.

S

S

S

A

A

S

A

u

A

u S

S

A

A

:

Figure 1.13 (a) A vector A lying in the xy plane can be represented by

Figure 1.14 The signs of the

its component vectors A x and A y. (b) The y component vector A y can be : moved to the right so that it adds to A x . The vector sum of the component : vectors is A . These three vectors form a right triangle.

components of a vector A depend on the quadrant in which the vector is located.

:

:

:

:

18  CHAPTER 1 | Introduction and Vectors :

S

B

Figure 1.15 The component vectors : of vector B in a coordinate system that is tilted.

Suppose a vector B makes an angle 9 with the x9 axis defined in Figure 1.15. The com: ponents of B along these axes are given by Bx9 5 B cos 9 and By9 5 B sin 9, as in Equa: tion 1.10. The magnitude and direction of B are obtained from expressions equivalent to Equations 1.11 and 1.12. Therefore, we can express the components of a vector in any coordinate system that is convenient for a particular situation. QUI C K QU IZ 1.6 Choose the correct response to make the sentence true: A component of a vector is (a) always, (b) never, or (c) sometimes larger than the magnitude of the vector.

Unit Vectors Vector quantities often are expressed in terms of unit vectors. A unit vector is a dimensionless vector having a magnitude of exactly 1. Unit vectors are used to specify a given direction and have no other physical significance. We shall use the symbols ˆi , ˆj , and k ˆ to represent unit vectors pointing in the x, y, and z directions, respectively. The “hat” over the letters is a common notation for a unit vector; for example, ˆi is ˆ form a set of mutually perpendicular vectors called “i-hat.” The unit vectors ˆi , ˆj , and k as shown in Active Figure 1.16a, where the magnitude of each unit vector equals 1; that is, u ˆi u 5 u ˆj u 5 u ˆk u 5 1. : Consider a vector A lying in the xy plane, as in Active Figure 1.16b. The product of : the component Ax and the unit vector ˆi is the component vector Ax 5 Axˆi , which lies on the x axis and has magnitude Ax . Likewise, Ay ˆj is a component vector of magnitude : Ay lying on the y axis. Therefore, the unit-vector notation for the vector A is : A 5 A ˆi 1 A ˆj 1.13b x

y

:

:

:

Now suppose we wish to add vector B to vector A, where B has components Bx and B y . The procedure for performing this sum is simply to add the x and y components : : : separately. The resultant vector R 5 A 1 B is therefore :

R 5 (Ax 1 Bx)ˆi 1 (Ay 1 By)ˆj

1.14b

From this equation, the components of the resultant vector are given by R x 5 Ax 1 B x

1.15b

Ry 5 Ay 1 By Therefore, we see that in the component method of adding vectors, we add all the x components to find the x component of the resultant vector and use the same

Active Figure 1.16 (a) The unit vectors ˆi , ˆj , and kˆ are directed along the x, y, and z axes, respectively. : (b) A vector A lying in the xy plane has component vectors Axˆi and Ay ˆj , where Ax and Ay are the components : of A . S

A

1.9 | Components of a Vector and Unit Vectors 19

process for the y components.  The procedure just described for adding two vec: : tors A and B using the component method can be checked using a diagram like Figure 1.17. : The magnitude of R and the angle it makes with the x axis can be obtained from its components using the relationships R 5 "Rx2 1 Ry2 5 "(Ax 1 Bx)2 1 (Ay 1 By)2 tan  5

Ry Rx

5

Ay 1 B y

1.16b

S

B

S

A

1.17b

Ax 1 Bx

The extension of these methods to three-dimensional vectors is straightforward. If : A and B both have x, y, and z components, we express them in the form

:

:

ˆ A 5 Ax ˆi 1 Ay ˆj 1 Az k : ˆ B 5 B ˆi 1 B ˆj 1 B k x

:

S

R

y

z

Figure 1.17 A geometric construction showing the relation between the components of the : resultant R of two vectors and the individual components.

:

The sum of A and B is : : : ˆ R 5 A 1 B 5 (Ax 1 Bx )ˆi 1 (Ay 1 By )ˆj 1 (Az 1 Bz )k

Pitfall Prevention | 1.8

1.18b

:

If a vector R has x, y, and z components, the magnitude of the vector is R 5 "Rx2 1 Ry2 1 Rz2 :

The angle x that R makes with the x axis is given by cos x 5

Rx R

with similar expressions for the angles with respect to the y and z axes. The extension of our method to adding more than two vectors is also : : : straightforward. For example, A 1 B 1 C 5 (Ax 1 Bx 1 Cx)ˆi 1 (Ay 1 By 1 Cy)jˆ 1 ˆ (Az 1 Bz 1 Cz)k. Adding displacement vectors is relatively easy to visualize. We can also add other types of vectors, such as velocity, force, and electric field vectors, which we will do in later chapters. QUICK QUIZ 1.7 If at least one component of a vector is a positive number, the vector cannot (a) have any component that is negative, (b) be zero, (c) have three dimensions. :

:

Q U I CK QUI Z 1.8 If A 1 B 5 0, the corresponding components of the two vectors : : A and B must be (a) equal, (b) positive, (c) negative, (d) of opposite sign.

THINKING PHYSICS 1.2

You may have asked someone directions to a destination in a city and been told something like, “Walk 3 blocks east and then 5 blocks south.” If so, are you experienced with vector components? Reasoning Yes, you are! Although you may not have thought of vector compo-

nent language when you heard these directions, that is exactly what the directions represent. The perpendicular streets of the city reflect an xy coordinate system; we can assign the x axis to the east–west streets, and the y axis to the north–south streets. Therefore, the comment of the person giving you directions can be translated as, “Undergo a displacement vector that has an x component of 13 blocks and a y component of 25 blocks.” You would arrive at the same destination by undergoing the y component first, followed by the x component, demonstrating the commutative law of addition. b

Tangents on Calculators Equation 1.17 involves the calculation of an angle by means of a tangent function. Generally, the inverse tangent function on calculators provides an angle between 2908 and 1908. As a consequence, if the vector you are studying lies in the second or third quadrant, the angle measured from the positive x axis will be the angle your calculator returns plus 1808.

20  CHAPTER 1 | Introduction and Vectors Example 1.6 | The

Sum of Two Vectors :

:

Find the sum of two displacement vectors A and B lying in the xy plane and given by :

: A 5 (2.0ˆi 1 2.0 ˆj) m and B 5 (2.0ˆi 2 4.0 ˆj) m

SOLUTION

: : Comparing this expression for A with the general expression A 5 Axˆi 1 Ay ˆj 1 Az kˆ, we see that Ax 5 2.0 m, Ay 5 2.0 m, and Az 5 0. Likewise, Bx 5 2.0 m, By 5 24.0 m, and Bz 5 0. We can use a two-dimensional approach because there are no z components. :

:

: : R 5 A 1 B 5 (2.0 1 2.0)ˆi m 1 (2.0 2 4.0) ˆj m

Use Equation 1.14 to obtain the resultant vector R: :

Evaluate the components of R:

Rx 5 4.0 m :

R y 5 22.0 m

R 5 "Rx2 1 Ry2 5 "(4.0 m)2 1 (22.0 m)2 5 "20 m 5 4.5 m

Use Equation 1.16 to find the magnitude of R: :

tan  5

Find the direction of R from Equation 1.17:

Ry Rx

5

22.0 m 5 20.50 4.0 m

Your calculator likely gives the answer 2278 for  5 tan21(20.50). This answer is correct if we interpret it to mean 278 clockwise from the x axis. Our standard form has been to quote the angles measured counterclockwise from the 1x axis, and that angle for this vector is  5 333.8

Example 1.7 | The

Resultant Displacement

A particle undergoes three consecutive displacements: D: r1 5 (15ˆi 1 30 ˆj 1 12kˆ) cm, D: r2 5 (23ˆi 2 14 ˆj 2 5.0kˆ) cm, and : ˆ ˆ D r3 5 (213 i 1 15 j ) cm. Find unit-vector notation for the resultant displacement and its magnitude.

SOLUTION Although x is sufficient to locate a point in one dimension, we need a vector : r to locate a point in two or three dimensions. The notation D : r is a generalization of the one-dimensional displacement Dx. Three-dimensional displacements are more difficult to conceptualize than those in two dimensions because the latter can be drawn on paper. For this problem, let us imagine that you start with your pencil at the origin of a piece of graph paper on which you have drawn x and y axes. Move your pencil 15 cm to the right along the x axis, then 30 cm upward along the y axis, and then 12 cm perpendicularly toward you away from the graph paper. This procedure provides the displacement described by D: r1. From this point, move your pencil 23 cm to the right parallel to the x axis, then 14 cm parallel to the graph paper in the 2y direction, and then 5.0 cm perpendicularly away from you toward the graph paper. You are now at the displacement from the origin described by D : r1 1 D : r2. From this point, move your pencil 13 cm to the left in the 2x direction, and (finally!) 15 cm parallel to the graph paper along the y axis. Your final position is at a displacement D : r1 1 D : r2 1 D : r3 from the origin. To find the resultant displacement, add the three vectors:

Find the magnitude of the resultant vector:

D: r 5 D: r1 1 D : r2 1 D: r3 5 (15 1 23 2 13)ˆi cm 1 (30 2 14 1 15)ˆj cm 1 (12 2 5.0 1 0)kˆ cm 5 (25ˆi 1 31 ˆj 1 7.0kˆ) cm R 5 "R x2 1 R y2 1 R z2 5 "(25 cm)2 1 (31 cm)2 1 (7.0 cm)2 5 40 cm

1.9 | Components of a Vector and Unit Vectors 21

Exampl e 1.8 | Taking

a Hike

A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the second day, she walks 40.0 km in a direction 60.08 north of east, at which point she discovers a forest ranger’s tower. (A) Determine the components of the hiker’s displacement for each day.

S S

SOLUTION :

If we denote the displacement vectors on the first and second days by A and : B, respectively, and use the car as the origin of coordinates, we obtain the vectors : shown in Figure 1.18. Drawing the resultant R, we see that this problem is one we’ve solved before: an addition of two vectors. :

Displacement A has a magnitude of 25.0 km and is directed 45.08 below the positive x axis. :

S

Figure 1.18 (Example 1.8) The total displacement of the hiker is the vector : : R 5 A 1 B. :

Ax 5 A cos (245.08) 5 (25.0 km)(0.707) 5 17.7 km

Find the components of A using Equation 1.10:

Ay 5 A sin (245.08) 5 (25.0 km)(20.707) 5 217.7 km The negative value of Ay indicates that the hiker walks in the negative y direction on the first day. The signs of Ax and Ay also are evident from Figure 1.18. :

Bx 5 B cos 60.08 5 (40.0 km)(0.500) 5 20.0 km

Find the components of B using Equation 1.10:

By 5 B sin 60.08 5 (40.0 km)(0.866) 5 34.6 km :

:

(B) Determine the components of the hiker’s resultant displacement R for the trip. Find an expression for R in terms of unit vectors.

SOLUTION Use Equation 1.15 to find the components of the resultant : : : displacement R 5 A 1 B:

Rx 5 Ax 1 Bx 5 17.7 km 1 20.0 km 5 37.7 km R y 5 Ay 1 By 5 217.7 km 1 34.6 km 5 17.0 km :

R 5 (37.7ˆi 1 17.0 ˆj ) km

Write the total displacement in unit-vector form:

Looking at the graphical representation in Figure 1.18, we estimate the position of the tower to be about (38 km, 17 km), : which is consistent with the components of R in our result for the final position of the hiker. Also, both components : of R are positive, putting the final position in the first quadrant of the coordinate system, which is also consistent with Figure 1.18.

What If? After reaching the tower, the hiker wishes to return to her car along a single straight line. What are the components of the vector representing this hike? What should the direction of the hike be? :

:

Answer The desired vector Rcar is the negative of vector R: :

:

Rcar 5 2R 5 (237.7ˆi 2 17.0 ˆj ) km The direction is found by calculating the angle that the vector makes with the x axis: tan  5

R car,y R car,x

which gives an angle of  5 204.28, or 24.28 south of west.

5

217.0 km 5 0.450 237.7 km

22  CHAPTER 1 | Introduction and Vectors

1.10 | Modeling, Alternative Representations, and Problem-Solving Strategy Most courses in general physics require the student to learn the skills of problem solving, and examinations usually include problems that test such skills. This section describes some useful ideas that will enable you to enhance your understanding of physical concepts, increase your accuracy in solving problems, eliminate initial panic or lack of direction in approaching a problem, and organize your work. One of the primary problem-solving methods in physics is to form an appropriate model of the problem. A model is a simplified substitute for the real problem that allows us to solve the problem in a relatively simple way. As long as the predictions of the model agree to our satisfaction with the actual behavior of the real system, the model is valid. If the predictions do not agree, the model must be refined or replaced with another model. The power of modeling is in its ability to reduce a wide variety of very complex problems to a limited number of classes of problems that can be approached in similar ways. In science, a model is very different from, for example, an architect’s scale model of a proposed building, which appears as a smaller version of what it represents. A scientific model is a theoretical construct and may have no visual similarity to the physical problem. A simple application of modeling is presented in Example 1.9, and we shall encounter many more examples of models as the text progresses. Models are needed because the actual operation of the Universe is extremely complicated. Suppose, for example, we are asked to solve a problem about the Earth’s motion around the Sun. The Earth is very complicated, with many processes occurring simultaneously. These processes include weather, seismic activity, and ocean movements as well as the multitude of processes involving human activity. Trying to maintain knowledge and understanding of all these processes is an impossible task. The modeling approach recognizes that none of these processes affects the motion of the Earth around the Sun to a measurable degree. Therefore, these details are all ignored. In addition, as we shall find in Chapter 11, the size of the Earth does not affect the gravitational force between the Earth and the Sun; only the masses of the Earth and Sun and the distance between them determine this force. In a simplified model, the Earth is imagined to be a particle, an object with mass but zero size. This replacement of an extended object by a particle is called the particle model, which is used extensively in physics. By analyzing the motion of a particle with the mass of the Earth in orbit around the Sun, we find that the predictions of the particle’s motion are in excellent agreement with the actual motion of the Earth. The two primary conditions for using the particle model are as follows: • The size of the actual object is of no consequence in the analysis of its motion. • Any internal processes occurring in the object are of no consequence in the analysis of its motion. Both of these conditions are in action in modeling the Earth as a particle. Its radius is not a factor in determining its motion, and internal processes such as thunderstorms, earthquakes, and manufacturing processes can be ignored. Four categories of models used in this book will help us understand and solve physics problems. The first category is the geometric model. In this model, we form a geometric construction that represents the real situation. We then set aside the real problem and perform an analysis of the geometric construction. Consider a popular problem in elementary trigonometry, as in the following example.

1.10 | Modeling, Alternative Representations, and Problem-Solving Strategy  23

Exampl e 1.9 | Finding

the Height of a Tree

You wish to find the height of a tree but cannot measure it directly. You stand 50.0 m from the tree and determine that a line of sight from the ground to the top of the tree makes an angle of 25.08 with the ground. How tall is the tree?

SOLUTION Figure 1.19 shows the tree and a right triangle corresponding to the information in the problem superimposed over it. (We assume that the tree is exactly perpendicular to a perfectly flat ground.) In the triangle, we know the length of the horizontal leg and the angle between the hypotenuse and the horizontal leg. We can find the height of the tree by calculating the length of the vertical leg. We do so with the tangent function: tan  5

opposite side adjacent side

5

h 50.0 m

h 5 (50.0 m) tan  5 (50.0 m) tan 25.08 5 23.3 m

Figure 1.19 (Example 1.9) The height of a tree can be found by measuring the distance from the tree and the angle of sight to the top above the ground. This problem is a simple example of geometrically modeling the actual problem.

You may have solved a problem very similar to Example 1.9 but never thought about the notion of modeling. From the modeling approach, however, once we draw the triangle in Figure 1.19, the triangle is a geometric model of the real problem; it is a substitute. Until we reach the end of the problem, we do not imagine the problem to be about a tree but to be about a triangle. We use trigonometry to find the vertical leg of the triangle, leading to a value of 23.3 m. Because this leg represents the height of the tree, we can now return to the original problem and claim that the height of the tree is 23.3 m. Other examples of geometric models include modeling the Earth as a perfect sphere, a pizza as a perfect disk, a meter stick as a long rod with no thickness, and an electric wire as a long, straight cylinder. The particle model is an example of the second category of models, which we will call simplification models. In a simplification model, details that are not significant in determining the outcome of the problem are ignored. When we study rotation in Chapter 10, objects will be modeled as rigid objects. All the molecules in a rigid object maintain their exact positions with respect to one another. We adopt this simplification model because a spinning rock is much easier to analyze than a spinning block of gelatin, which is not a rigid object. Other simplification models will assume that quantities such as friction forces are negligible, remain constant, or are proportional to some power of the object’s speed. The third category is that of analysis models, which are general types of problems that we have solved before. An important technique in problem solving is to cast a new problem into a form similar to one we have already solved and which can be used as a model. As we shall see, there are about two dozen analysis models that can be used to solve most of the problems you will encounter. We will see our first analysis models in Chapter 2, where we will discuss them in more detail. The fourth category of models is structural models. These models are generally used to understand the behavior of a system that is far different in scale from our macroscopic world — either much smaller or much larger — so that we cannot interact with it directly. As an example, the notion of a hydrogen atom as an electron in a circular orbit around a proton is a structural model of the atom. We will discuss this model and structural models in general in Chapter 11.

24  CHAPTER 1 | Introduction and Vectors Intimately related to the notion of modeling is that of forming alternative representations of the problem. A representation is a method of viewing or presenting the information related to the problem. Scientists must be able to communicate complex ideas to individuals without scientific backgrounds. The best representation to use in conveying the information successfully will vary from one individual to the next. Some will be convinced by a well-drawn graph, and others will require a picture. Physicists are often persuaded to agree with a point of view by examining an equation, but nonphysicists may not be convinced by this mathematical representation of the information. A word problem, such as those at the ends of the chapters in this book, is one representation of a problem. In the “real world” that you will enter after graduation, the initial representation of a problem may be just an existing situation, such as the effects of global warming or a patient in danger of dying. You may have to identify the important data and information, and then cast the situation yourself into an equivalent word problem! Considering alternative representations can help you think about the information in the problem in several different ways to help you understand and solve it. Several types of representations can be of assistance in this endeavor:

Figure 1.20 A pictorial representation of a pop foul being hit by a baseball player.

S

Figure 1.21 A simplified pictorial representation for the situation shown in Figure 1.20.

Figure 1.22 A graphical representation of the position as a function of time of a block hanging from a spring and oscillating.

• Mental representation. From the description of the problem, imagine a scene that describes what is happening in the word problem, then let time progress so that you understand the situation and can predict what changes will occur in the situation. This step is critical in approaching every problem. • Pictorial representation. Drawing a picture of the situation described in the word problem can be of great assistance in understanding the problem. In Example 1.9, the pictorial representation in Figure 1.19 allows us to identify the triangle as a geometric model of the problem. In architecture, a blueprint is a pictorial representation of a proposed building. Generally, a pictorial representation describes what you would see if you were observing the situation in the problem. For example, Figure 1.20 shows a pictorial representation of a baseball player hitting a short pop foul. Any coordinate axes included in your pictorial representation will be in two dimensions: x and y axes. • Simplified pictorial representation. It is often useful to redraw the pictorial representation without complicating details by applying a simplification model. This process is similar to the discussion of the particle model described earlier. In a pictorial representation of the Earth in orbit around the Sun, you might draw the Earth and the Sun as spheres, with possibly some attempt to draw continents to identify which sphere is the Earth. In the simplified pictorial representation, the Earth and the Sun would be drawn simply as dots, representing particles. Figure 1.21 shows a simplified pictorial representation corresponding to the pictorial representation of the baseball trajectory in Figure 1.20. The notations vx and vy refer to the components of the velocity vector for the baseball. We shall use such simplified pictorial representations throughout the book. • Graphical representation. In some problems, drawing a graph that describes the situation can be very helpful. In mechanics, for example, position – time graphs can be of great assistance. Similarly, in thermodynamics, pressure – volume graphs are essential to understanding. Figure 1.22 shows a graphical representation of the position as a function of time of a block on the end of a vertical spring as it oscillates up and down. Such a graph is helpful for understanding simple harmonic motion, which we study in Chapter 12. A graphical representation is different from a pictorial representation, which is also a two-dimensional display of information but whose axes, if any, represent length coordinates. In a graphical representation, the axes may represent any two related variables. For example, a graphical representation may have axes for temperature and time. Therefore, in comparison to a pictorial representation, a

1.10 | Modeling, Alternative Representations, and Problem-Solving Strategy  25

graphical representation is generally not something you would see when observing the situation in the problem with your eyes. • Tabular representation. It is sometimes helpful to organize the information in tabular form to help make it clearer. For example, some students find that making tables of known quantities and unknown quantities is helpful. The periodic table of the elements is an extremely useful tabular representation of information in chemistry and physics. • Mathematical representation. The ultimate goal in solving a problem is often the mathematical representation. You want to move from the information contained in the word problem, through various representations of the problem that allow you to understand what is happening, to one or more equations that represent the situation in the problem and that can be solved mathematically for the desired result. Besides what you might expect to learn about physics concepts, a very valuable skill you should acquire from your physics course is the ability to solve complicated problems. The way physicists approach complex situations and break them into manageable pieces is extremely useful. The following is a general problem-solving strategy to guide you through the steps. To help you remember the steps of the strategy, they are Conceptualize, Categorize, Analyze, and Finalize.

G E N E RAL PRO B L E M -SO LVIN G ST RAT EGY Conceptualize

• The first things to do when approaching a problem are to think about and understand the situation. Study carefully any representations of the information (for example, diagrams, graphs, tables, or photographs) that accompany the problem. Imagine a movie, running in your mind, of what happens in the problem. • If a pictorial representation is not provided, you should almost always make a quick drawing of the situation. Indicate any known values, perhaps in a table or directly on your sketch. • Now focus on what algebraic or numerical information is given in the problem. Carefully read the problem statement, looking for key phrases such as “starts from rest” (vi 5 0) or “stops” (vf 5 0). • Now focus on the expected result of solving the problem. Exactly what is the question asking? Will the final result be numerical or algebraic? Do you know what units to expect? • Don’t forget to incorporate information from your own experiences and common sense. What should a reasonable answer look like? For example, you wouldn’t expect to calculate the speed of an automobile to be 5 3 106 m/s. Categorize

• Once you have a good idea of what the problem is about, you need to simplify the problem. Remove the details that are not important to the solution. For example, model a moving object as a particle. If appropriate, ignore air resistance or friction between a sliding object and a surface. • Once the problem is simplified, it is important to categorize the problem. Is it a simple substitution problem such that numbers can be substituted into an equation? If so, the problem is likely to be finished when this substitution is done. If not, you face what we call an analysis problem: the situation must be analyzed more deeply to reach a solution. • If it is an analysis problem, it needs to be categorized further. Have you seen this type of problem before? Does it fall into the growing list of types of problems that you have solved previously? If so, identify any analysis

26  CHAPTER 1 | Introduction and Vectors model(s) appropriate for the problem to prepare for the Analyze step below. Being able to classify a problem with an analysis model can make it much easier to lay out a plan to solve it. For example, if your simplification shows that the problem can be treated as a particle under constant acceleration and you have already solved such a problem (such as the examples we shall see in Section 2.6), the solution to the present problem follows a similar pattern. Analyze

• Now you must analyze the problem and strive for a mathematical solution. Because you have already categorized the problem and identified an analysis model, it should not be too difficult to select relevant equations that apply to the type of situation in the problem. For example, if the problem involves a particle under constant acceleration (which we will study in Section 2.6), Equations 2.10 to 2.14 are relevant. • Use algebra (and calculus, if necessary) to solve symbolically for the unknown variable in terms of what is given. Substitute in the appropriate numbers, calculate the result, and round it to the proper number of significant figures. Finalize

• Examine your numerical answer. Does it have the correct units? Does it meet your expectations from your conceptualization of the problem? What about the algebraic form of the result? Does it make sense? Examine the variables in the problem to see whether the answer would change in a physically meaningful way if the variables were drastically increased or decreased or even became zero. Looking at limiting cases to see whether they yield expected values is a very useful way to make sure that you are obtaining reasonable results. • Think about how this problem compared with others you have solved. How was it similar? In what critical ways did it differ? Why was this problem assigned? Can you figure out what you have learned by doing it? If it is a new category of problem, be sure you understand it so that you can use it as a model for solving similar problems in the future. When solving complex problems, you may need to identify a series of subproblems and apply the problem-solving strategy to each. For simple problems, you probably don’t need this strategy. When you are trying to solve a problem and you don’t know what to do next, however, remember the steps in the strategy and use them as a guide. In the rest of this book, we will label the Conceptualize, Categorize, Analyze, and Finalize steps explicitly in the worked examples. Many chapters in this book include a section labeled Problem-Solving Strategy that should help you through the rough spots. These sections are organized according to the General Problem-Solving Strategy outlined above and are tailored to the specific types of problems addressed in that chapter. To clarify how this Strategy works, we repeat Example 1.8 on the next page with the particular steps of the Strategy identified.

1.10 | Modeling, Alternative Representations, and Problem-Solving Strategy  27

Simplify the problem. Remove the details that are not important to the solution. Then Categorize the problem. Is it a simple substitution problem such that numbers can be substituted into an equation? If not, you face an analysis problem. In this case, identify the appropriate analysis model. (Analysis models will be introduced in Chapter 2.)

When you Conceptualize a problem, try to understand the situation that is presented in the problem statement. Study carefully any representations of the information (for example, diagrams, graphs, tables, or photographs) that accompany the problem. Imagine a movie, running in your mind, of what happens in the problem.

Exampl m e 1.8 | Taking

Now Analyze the problem. Select relevant equations from the analysis model. Solve symbolically for the unknown variable in terms of what is given. Substitute in the appropriate numbers, calculate the result, and round it to the proper number of significant figures.

a Hike

A hikerr begins a trip by first walking 25.0 km southeast from her car. She stop stops and sets up her h tent for the night. On the second day, she walks 40.0 km in a direction 60.08 north of east, at which point tower. o int she discovers a forest ranger’s tow (A) Determine the components for each day. te nen of the hiker’s displacement nt fo

S S

SOLUTION I Conceptualize We conceptualize c the problem m by drawing a sketch as in Figure 1.18. : : If we denote the vectors on the first and second days by A and B, rehe displacement d spectively, and nd use the car as the origin n of coordinates, we obtain the vectors shown in Figure 1.1 1.18.

S

:

Categorize Drawing the resultant esult R, we can now categorize this problem as one

Figure 1.18 (Example 1.8) The total

we’ve solved before: an add addition of two vectors. You should now have a hint of the displacement of the hiker is the vector : : : power of categorization tion in that many new problems are very similar to problems we R 5 A 1 B. have already solved ved if we are careful to conceptualize them. Once we have drawn the displacement vectors and categorized the problem, this problem is no longer about a hiker, a walk, a car, a tent, or a tower. nt ve It is a problem oble about vector addition, one that we have already solved. :

Analyze Displacement A has a magnitude of 25.0 km and is directed 45.08 below the positive x axis. :

Find the components of A using Equation 1.10:

Ax 5 A cos(245.08) 5 (25.0 km)(0.707) 5 17.7 km Ay 5 A sin(245.08) 5 (25.0 km)(20.707) 5 217.7 km

The negative value of Ay indicates that the hiker walks in the negative y direction on the first day. The signs of Ax and Ay also are evident from Figure 1.18. :

Find the components of B using Equation 1.10:

Bx 5 B cos 60.08 5 (40.0 km)(0.500) 5 20.0 km By 5 B sin 60.08 5 (40.0 km)(0.866) 5 34.6 km :

:

(B) Determine the components of the hiker’s resultant displacement R for the trip. Find an expression for R in terms of unit vectors.

SOLUTION Use Equation 1.15 to find the components of the resultant : : : displacement R 5 A 1 B: Write the total displacement in unit-vector form:

Rx 5 Ax 1 Bx 5 17.7 km 1 20.0 km 5 37.7 km R y 5 Ay 1 By 5 217.7 km 1 34.6 km 5 17.0 km :

R 5 (37.7ˆi 1 17.0 ˆj ) km

28  CHAPTER 1 | Introduction and Vectors

Finalize the problem. Examine the numerical answer. Does it have the correct units? Does it meet your expectations from your conceptualization of the problem? Does the answer make sense? What about the algebraic form of the result? Examine the variables in the problem to see whether the answer would change in a physically meaningful way if the variables were drastically increased or decreased or even became zero.

What If? questions will appear in many examples in the text, and offer a variation on the situation just explored. This feature encourages you to think about the results of the example and assists in conceptual understanding of the principles.

1.8 cont. Finalize Looking at the graphical graph representation in Figure 1.18, we estimate the position of the tower to be about (38 km, : 17 km), which is consistent nsiste with the components of R in our result for the final position of the hiker. Also, both compo: nents of R are positive, positi putting the final position in the first quadrant of the coordinate system, which is also consistent with Figure 1.18. 8.

What If? After reaching the tower, the hiker wishes to return to her car along a single straight line. What are the components of the vector representing this hike? What should the direction of the hike be? :

:

Answer The desired vector Rcar is the negative of vector R: :

:

Rcar 5 2R 5 (237.7ˆi 2 17.0 ˆj ) km The direction is found by calculating the angle that the vector makes with the x axis: tan  5

R car,y R car,x

5

217.0 km 5 0.450 237.7 km

which gives an angle of  5 204.28, or 24.28 south of west.

SUMMARY | Mechanical quantities can be expressed in terms of three fundamental quantities—length, mass, and time—which in the SI system have the units meters (m), kilograms (kg), and seconds (s), respectively. It is often useful to use the method of dimensional analysis to check equations and to assist in deriving expressions. The density of a substance is defined as its mass per unit volume: ;

m V

1.1b

Vectors are quantities that have both magnitude and direction and obey the vector law of addition. Scalars are quantities that add algebraically. : : Two vectors A and B can be added using the triangle method. In this method (see Active Fig. 1.9), the vector : : : : : R 5 A 1 B runs from the tail of A to the tip of B. : The x component Ax of the vector A is equal to its projection along the x axis of a coordinate system, where Ax 5 A cos  : and where  is the angle A makes with the x axis. Likewise, : the y component Ay of A is its projection along the y axis, where Ay 5 A sin . : If a vector A has an x component equal to Ax and a y component equal to Ay , the vector can be expressed in

:

unit-vector form as A 5 (Axˆi 1 Ay ˆj ). In this notation, ˆi is a unit vector in the positive x direction and ˆj is a unit vector in the positive y direction. Because ˆi and ˆj are unit vectors, |ˆi | 5 | ˆj | 5 1. In three dimensions, a vector can be expressed : ˆ), where k ˆ is a unit vector in the as A 5 (Axˆi 1 Ay ˆj 1 Az k z direction. The resultant of two or more vectors can be found by resolving all vectors into their x, y, and z components and adding their components: :

:

:

R 5 A 1 B 5 (Ax 1 Bx)ˆi 1 (Ay 1 By) ˆj 1 (Az 1 Bz)kˆ

1.18b

Problem-solving skills and physical understanding can be improved by modeling the problem and by constructing alternative representations of the problem. Models helpful in solving problems include geometric, simplification, and analysis models. Scientists use structural models to understand systems larger or smaller in scale than those with which we normally have direct experience. Helpful representations include the mental, pictorial, simplified pictorial, graphical, tabular, and mathematical representations. Complicated problems are best approached in an organized manner. Recall and apply the Conceptualize, Categorize, Analyze, and Finalize steps of the General Problem-Solving Strategy when you need them.

| Objective Questions  29 denotes answer available in Student Solutions Manual/Study Guide

OBJECTIVE QUESTIONS | 1. Answer each question yes or no. Must two quantities have the same dimensions (a) if you are adding them? (b) If you are multiplying them? (c) If you are subtracting them? (d) If you are dividing them? (e) If you are equating them? 2. The price of gasoline at a particular station is 1.5 euros per liter. An American student can use 33 euros to buy gasoline. Knowing that 4 quarts make a gallon and that 1 liter is close to 1 quart, she quickly reasons that she can buy how many gallons of gasoline? (a) less than 1 gallon (b) about 5 gallons (c) about 8 gallons (d) more than 10 gallons 3. Rank the following five quantities in order from the largest to the smallest. If two of the quantities are equal, give them equal rank in your list. (a) 0.032 kg (b) 15 g (c) 2.7 3 105 mg (d) 4.1 3 1028 Gg (e) 2.7 3 108 g ˆ) m/s? 4. What is the y component of the vector (3ˆi 2 8k (a) 3 m/s (b) 28 m/s (c) 0 (d) 8 m/s (e) none of those answers 5. Which of the following is the best estimate for the mass of all the people living on the Earth? (a) 2 3 108 kg (b) 1 3 109 kg (c) 2 3 1010 kg (d) 3 3 1011 kg (e) 4 3 1012 kg

10. What is the y component of the vector shown in Figure OQ1.9? (a) 3 cm (b) 6 cm (c) 24 cm (d) 26 cm (e) none of those answers 11. Yes or no: Is each of the following quantities a vector? (a) force (b) temperature (c) the volume of water in a can (d) the ratings of a TV show (e) the height of a building (f) the velocity of a sports car (g) the age of the Universe 12. A vector lying in the xy plane has components of opposite sign. The vector must lie in which quadrant? (a) the first quadrant (b) the second quadrant (c) the third quadrant (d) the fourth quadrant (e) either the second or the fourth quadrant :

:

13. Figure OQ1.13 shows two vectors D1 and D2. Which of the : : possibilities (a) through (d) is the vector D2 2 2D1, or (e) is it none of them? S

D S

D

6. What is the sum of the measured values 21.4 s 1 15 s 1 17.17 s 1 4.00 3 s? (a) 57.573 s (b) 57.57 s (c) 57.6 s (d) 58 s (e) 60 s 7. One student uses a meterstick to measure the thickness of a textbook and obtains 4.3 cm 6 0.1 cm. Other students measure the thickness with vernier calipers and obtain four different measurements: (a) 4.32 cm 6 0.01 cm, (b) 4.31 cm 6 0.01 cm, (c) 4.24 cm 6 0.01 cm, and (d) 4.43 cm 6 0.01 cm. Which of these four measurements, if any, agree with that obtained by the first student? 8. Newton’s second law of motion (Chapter 4) says that the mass of an object times its acceleration is equal to the net force on the object. Which of the following gives the correct units for force? (a) kg ? m/s2 (b) kg ? m2/s2 (c) kg/m ? s2 (d) kg ? m2/s (e) none of those answers 9. What is the x component of the vector shown in Figure OQ1.9? (a) 3 cm (b) 6 cm (c) 24 cm (d) 26 cm (e) none of those answers

Figure OQ1.13

14. A vector points from the origin into the second quadrant of the xy plane. What can you conclude about its components? (a) Both components are positive. (b) The x component is positive, and the y component is negative. (c) The x component is negative, and the y component is positive. (d) Both components are negative. (e) More than one answer is possible. ˆ) m/s? 15. What is the magnitude of the vector (10ˆi 2 10 k (a) 0 (b) 10 m/s (c) 210 m/s (d) 10 (e) 14.1 m/s :

16. Vector A lies in the xy plane. Both of its components will be negative if it points from the origin into which quadrant? (a) the first quadrant (b) the second quadrant (c) the third quadrant (d) the fourth quadrant (e) the second or fourth quadrants

Figure OQ1.9 Objective Questions 9 and 10.

CONCEPTUAL QUESTIONS |

denotes answer available in Student Solutions Manual/Study Guide

1. What natural phenomena could serve as alternative time standards?

would you need to address to make sure that the standard of density is as accurate as possible?

2. Express the following quantities using the prefixes given in Table 1.4. (a) 3 3 1024 m (b) 5 3 1025 s (c) 72 3 102 g

4. If the component of vector A along the direction of vector : B is zero, what can you conclude about the two vectors?

3. Suppose the three fundamental standards of the metric system were length, density, and time rather than length, mass, and time. The standard of density in this system is to be defined as that of water. What considerations about water

5. A book is moved once around the perimeter of a tabletop with the dimensions 1.0 m by 2.0 m. The book ends up at its initial position. (a) What is its displacement? (b) What is the distance traveled?

:

30  CHAPTER 1 | Introduction and Vectors 6. Can the magnitude of a vector have a negative value? Explain.

giving its angle measured counterclockwise from the positive x axis? In what cases will it be incorrect?

7. On a certain calculator, the inverse tangent function returns a value between –908 and 1908. In what cases will this value correctly state the direction of a vector in the xy plane, by

8. Is it possible to add a vector quantity to a scalar quantity? Explain.

PROBLEMS | The problems found in this chapter may be assigned online in Enhanced WebAssign.

denotes Master It tutorial available in Enhanced WebAssign denotes asking for quantitative and conceptual reasoning denotes symbolic reasoning problem denotes “paired problems” that develop reasoning with symbols and numerical values denotes Watch It video solution available in Enhanced WebAssign

1. denotes straightforward problem; 2. denotes intermediate problem; 3. denotes challenging problem 1. denotes full solution available in the Student Solutions Manual/ Study Guide

1. denotes problems most often assigned in Enhanced WebAssign. denotes biomedical problem denotes guided problem

quantity it describes: (d) the total circumference of the flat circular faces, (e) the volume, or (f) the area of the curved surface.

Section 1.1 Standards of Length, Mass, and Time Note: Consult the endpapers, appendices, and tables in the text whenever necessary in solving problems. For this chapter, Appendix B.3 and Table 15.1 may be particularly useful. Answers to odd-numbered problems appear in the back of the book. 1.

Two spheres are cut from a certain uniform rock. One has radius 4.50 cm. The mass of the other is five times greater. Find its radius.

2.

(a) Use information on the endpapers of this book to calculate the average density of the Earth. (b) Where does the value fit among those listed in Table 15.1 in Chapter 15? Look up the density of a typical surface rock like granite in another source and compare it with the density of the Earth.

3. An automobile company displays a die-cast model of its first car, made from 9.35 kg of iron. To celebrate its hundredth year in business, a worker will recast the model in gold from the original dies. What mass of gold is needed to make the new model? 4.

What mass of a material with density  is required to make a hollow spherical shell having inner radius r 1 and outer radius r 2?

Section 1.2 Dimensional Analysis

Figure P1.6 shows a frustum of a cone. Match each of the three expressions (a) (r1 1 r2)[h2 1 (r2 2 r1)2]1/2, (b) 2(r1 1 r2), and (c) h(r12 1 r1r2 1 r22)y3 with the

Section 1.3 Conversion of Units 8. A section of land has an area of 1 square mile and contains 640 acres. Determine the number of square meters in 1 acre. 9.

One gallon of paint (volume 5 3.78 3 10–3 m3) covers an area of 25.0 m2. What is the thickness of the fresh paint on the wall?

10.

Suppose your hair grows at the rate 1/32 in. per day. Find the rate at which it grows in nanometers per second. Because the distance between atoms in a molecule is on the order of 0.1 nm, your answer suggests how rapidly layers of atoms are assembled in this protein synthesis.

11. An ore loader moves 1 200 tons/h from a mine to the surface. Convert this rate to pounds per second, using 1 ton 5 2 000 lb. 12. The mass of the Sun is 1.99 3 1030 kg, and the mass of an atom of hydrogen, of which the Sun is mostly composed, is 1.67 3 10227 kg. How many atoms are in the Sun?

5. Which of the following equations are dimensionally correct? (a) vf 5 vi 1 ax (b) y 5 (2 m) cos (kx), where k 5 2 m21 6.

7. The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position as x 5 kamt n, where k is a dimensionless constant. Show by dimensional analysis that this expression is satisfied if m 5 1 and n 5 2. Can this analysis give the value of k?

13.

Figure P1.6

One cubic meter (1.00 m3) of aluminum has a mass of 2.70 3 103 kg, and the same volume of iron has a mass of 7.86 3 103 kg. Find the radius of a solid aluminum sphere that will balance a solid iron sphere of radius 2.00 cm on an equal-arm balance.

| Problems  31 14.

15.

Let Al represent the density of aluminum and  Fe that of iron. Find the radius of a solid aluminum sphere that balances a solid iron sphere of radius r Fe on an equal-arm balance. Assume it takes 7.00 min to fill a 30.0-gal gasoline tank. (a) Calculate the rate at which the tank is filled in gallons per second. (b) Calculate the rate at which the tank is filled in cubic meters per second. (c) Determine the time interval, in hours, required to fill a 1.00-m3 volume at the same rate. (1 U.S. gal 5 231 in.3)

16. A hydrogen atom has a diameter of 1.06 3 10210 m. The nucleus of the hydrogen atom has a diameter of approximately 2.40 3 10215 m. (a) For a scale model, represent the diameter of the hydrogen atom by the playing length of an American football field (100 yards 5 300 ft) and determine the diameter of the nucleus in millimeters. (b) Find the ratio of the volume of the hydrogen atom to the volume of its nucleus.

Section 1.4 Order-of-Magnitude Calculations 17. Find the order of magnitude of the number of table-tennis balls that would fit into a typical-size room (without being crushed). 18. (a) Compute the order of magnitude of the mass of a bathtub half full of water. (b) Compute the order of magnitude of the mass of a bathtub half full of copper coins. 19. To an order of magnitude, how many piano tuners reside in New York City? The physicist Enrico Fermi was famous for asking questions like this one on oral Ph.D. qualifying examinations. 20. An automobile tire is rated to last for 50 000 miles. To an order of magnitude, through how many revolutions will it turn over its lifetime?

thick, what volume of concrete is needed and what is the approximate uncertainty of this volume? Note: The next four problems call upon mathematical skills that will be useful throughout the course. 26.

Review. From the set of equations p 5 3q pr 5 qs 1 2 2 pr

1

1 2 2 qs

5 12qt 2

involving the unknowns p, q, r, s, and t, find the value of the ratio of t to r. 27. Review. A highway curve forms a section of a circle. A car goes around the curve as shown in the helicopter view of Figure u P1.27. Its dashboard compass shows that the car is initially headFigure P1.27 ing due east. After it travels d 5 840 m, it is heading  5 35.08 south of east. Find the radius of curvature of its path. Suggestion: You may find it useful to learn a geometric theorem stated in Appendix B.3. 28. Review. Prove that one solution of the equation 2.00x4 2 3.00x3 1 5.00x 5 70.0 is x 5 22.22. 29. Review. Find every angle  between 0 and 3608 for which the ratio of sin  to cos  is 23.00.

Section 1.6 Coordinate Systems Section 1.5 Significant Figures 21. The tropical year, the time interval from one vernal equinox to the next vernal equinox, is the basis for our calendar. It contains 365.242 199 days. Find the number of seconds in a tropical year.

30. Two points in the xy plane have Cartesian coordinates (2.00, 24.00) m and (23.00, 3.00) m. Determine (a) the distance between these points and (b) their polar coordinates. 31.

The polar coordinates of a point are r 5 5.50 m and  5 2408. What are the Cartesian coordinates of this point?

22.

Carry out the arithmetic operations (a) the sum of the measured values 756, 37.2, 0.83, and 2; (b) the product 0.003 2 3 356.3; and (c) the product 5.620 3 .

32.

Let the polar coordinates of the point (x, y) be (r, ). Determine the polar coordinates for the points (a) (2x, y), (b) (22x, 22y), and (c) (3x, 23y).

23.

How many significant figures are in the following numbers? (a) 78.9 6 0.2 (b) 3.788 3 109 (c) 2.46 3 10–6 (d) 0.005 3

33.

A fly lands on one wall of a room. The lower-left corner of the wall is selected as the origin of a two-dimensional Cartesian coordinate system. If the fly is located at the point having coordinates (2.00, 1.00) m, (a) how far is it from the origin? (b) What is its location in polar S coordinates? B

Note: Appendix B.8 on propagation of uncertainty may be useful in solving the next two problems. 24. The radius of a uniform solid sphere is measured to be (6.50 6 0.20) cm, and its mass is measured to be (1.85 6 0.02) kg. Determine the density of the sphere in kilograms per cubic meter and the uncertainty in the density.

Section 1.7 Vectors and Scalars

25. A sidewalk is to be constructed around a swimming pool that measures (10.0 6 0.1) m by (17.0 6 0.1) m. If the sidewalk is to measure (1.00 6 0.01) m wide by (9.0 6 0.1) cm

34.

S

A

Section 1.8 Some Properties of Vectors

u :

The displacement vectors A : and B shown in Figure P1.34 have

Figure P1.34 Problems 34 and 48.

32  CHAPTER 1 | Introduction and Vectors :

:

magnitudes of 3.00 m. The direction of vector A is  5 : : : : : : 30.08. Find graphically (a) A 1 B , (b) A 2 B , (c) B 2 A , : : and (d) A 2 2B . (Report all angles counterclockwise from the positive x axis.) 35. Why is the following situation impossible? A skater glides along a circular path. She defines a certain point on the circle as her origin. Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin. 36. A plane flies from base camp to Lake A, 280 km away in the direction 20.08 north of east. After dropping off supplies, it flies to Lake B, which is 190 km at 30.08 west of north from Lake A. Graphically determine the distance and direction from Lake B to the base camp. 37. A roller-coaster car moves 200 ft horizontally and then rises 135 ft at an angle of 30.08 above the horizontal. It next travels 135 ft at an angle of 40.08 downward. What is its displacement from its starting point? Use graphical techniques.

Section 1.9 Components of a Vector and Unit Vectors 38.

39.

: : Given the vectors A 5 2.00ˆi 1 6.00ˆj and B 5 3.00 ˆi 2 : : : ˆ 2.00 j, (a) draw the vector sum C 5 A 1 B and the vector : : : : : difference D 5 A 2 B. (b) Calculate C and D, in terms of : : unit vectors. (c) Calculate C and D in terms of polar coordinates, with angles measured with respect to the positive x axis. : : (a) Taking A 5 (6.00ˆi 2 8.00 ˆj ) units, B 5 (28.00ˆi 1 : ˆ ˆ ˆ 3.00 j ) units, and C 5 (26.0 i 1 19.0 j ) units, determine a : : : and b such that aA 1 bB 1 C 5 0. (b) A student has learned that a single equation cannot be solved to determine values for more than one unknown in it. How would you explain to him that both a and b can be determined from the single equation used in part (a)?

40. Find the horizontal and vertical components of the 100-m displacement of a superhero who flies from the top of a tall building following the path shown in Figure P1.40. 41.

A vector has an x component of 225.0 units and a y component of 40.0 units. Find the magnitude and direction of this vector.

Figure P1.40

:

:

The vector A has x, y, and z components of 8.00, 12.0, and 24.00 units, respectively. (a) Write a vector expression : for A in unit-vector notation. (b) Obtain a unit-vector ex: : pression for a vector B one-fourth the length of A pointing : in the same direction as A . (c) Obtain a unit-vector expres: : sion for a vector C three times the length of A pointing in : the direction opposite the direction of A .

44. Three displacement vectors of a croquet ball are shown in : : Figure P1.44, where u Au 5 20.0 units, u Bu 5 40.0 units, and

S

B

S

A

45.

A man pushing a mop across a floor causes it to undergo two displacements. S The first has a magnitude of 150 cm and C makes an angle of 1208 with the positive x axis. The resultant displacement has a magnitude of 140 cm and is diFigure P1.44 rected at an angle of 35.08 to the positive x axis. Find the magnitude and direction of the second displacement.

46.

Vector A has x and y components of 28.70 cm and : 15.0 cm, respectively; vector B has x and y components of : : : 13.2 cm and 26.60 cm, respectively. If A 2 B 1 3C 5 0, : what are the components of C?

47.

:

: : Consider the two vectors A 5 3ˆi 2 2 ˆj and B 5 2iˆ 2 4ˆj . : : : : : : : : Calculate (a) A 1 B, (b) A 2 B, (c) u A 1 B u , (d) u A 2 B u , : : : : and (e) the directions of A 1 B and A 2 B. :

:

48. Use the component method to add the vectors A and B : : shown in Figure P1.34. Express the resultant A 1 B in unit-vector notation. 49. In an assembly operation illustrated in Figure P1.49, a robot moves an object first straight upward and then also to the east, around an arc forming one-quarter of a circle of radius 4.80 cm that lies in an east–west vertical plane. The robot then moves the object upward and to the north, through one-quarter of a circle of radius 3.70 cm Figure P1.49 that lies in a north–south vertical plane. Find (a) the magnitude of the total displacement of the object and (b) the angle the total displacement makes with the vertical. 50.

42. Vector B has x, y, and z components of 4.00, 6.00, and : 3.00 units, respectively. Calculate (a) the magnitude of B : and (b) the angle that B makes with each coordinate axis. 43.

u Cu 5 30.0 units. Find (a) the resultant in unit-vector notation and (b) the magnitude and direction of the resultant displacement.

51.

: Consider the three displacement vectors A 5 (3ˆi 2 : : ˆ ˆ ˆ ˆ ˆ 3 j ) m, B 5 ( i 2 4 j) m, and C 5 (2 i 1 5 j) m. Use the component method to determine (a) the magnitude and : : : : direction of the vector D 5 A 1 B 1 C and (b) the magni: : : : tude and direction of E 5 2A 2 B 1 C.

A person going for a walk follows the path shown in Figure P1.51. The total trip consists of four straight-line paths. At the end of the walk, what is the person’s resultant displacement measured from the starting point?

52. Express in unit-vector notation the following vectors, each of which has magnitude 17.0 cm. : (a) Vector E is directed 27.08 counterclockwise from the posiFigure P1.51 : tive x axis. (b) Vector F is directed 27.08 counterclockwise from the positive y axis. (c) : Vector G is directed 27.08 clockwise from the negative y axis.

| Problems  33 Find the order of magnitude of the number of stars in the Milky Way. Assume the distance between the Sun and our nearest neighbor is typical.

Section 1.10 Modeling, Alternative Representations, and Problem-Solving Strategy 53. A crystalline solid consists of atoms stacked up in a repeating lattice structure. Consider a crystal as shown in Figure P1.53a. The atoms reside at the corners of cubes of side L 5 0.200 nm. One piece of evidence for the regular arrangement of atoms comes from the flat surfaces along which a crystal separates, or cleaves, Figure P1.53 when it is broken. Suppose this crystal cleaves along a face diagonal as shown in Figure P1.53b. Calculate the spacing d between two adjacent atomic planes that separate when the crystal cleaves.

57. In a situation in which data are known to three significant digits, we write 6.379 m 5 6.38 m and 6.374 m 5 6.37 m. When a number ends in 5, we arbitrarily choose to write 6.375 m 5 6.38 m. We could equally well write 6.375 m 5 6.37 m, “rounding down” instead of “rounding up,” because we would change the number 6.375 by equal increments in both cases. Now consider an order-of-magnitude estimate, in which factors of change rather than increments are important. We write 500 m ,103 m because 500 differs from 100 by a factor of 5 while it differs from 1 000 by only a factor of 2. We write 437 m ,103 m and 305 m ,102 m. What distance differs from 100 m and from 1 000 m by equal factors so that we could equally well choose to represent its order of magnitude as ,102 m or as ,103 m? 58.

:

54. As she picks up her riders, a bus driver traverses four successive displacements represented by the expression

1 (3.00 b cos 508)ˆi 2(3.00 b sin 508)ˆj 2(5.00 b)ˆj

60. In physics, it is important to use mathematical approximations. (a) Demonstrate that for small angles (, 208)

Here b represents one city block, a convenient unit of distance of uniform size; ˆi is east; and ˆj is north. The displacements at 408 and 508 represent travel on roadways in the city that are at these angles to the main east–west and north– south streets. (a) Draw a map of the successive displacements. (b) What total distance did she travel? (c) Compute the magnitude and direction of her total displacement. The logical structure of this problem and of several problems in later chapters was suggested by Alan Van Heuvelen and David Maloney, American Journal of Physics 67(3) 252–256, March 1999. A surveyor measures the distance across a straight river by the following method (Fig. P1.55). Starting directly across from a tree on the opposite bank, she walks d 5 100 m along the riverbank to establish a baseline. Then she sights across to the tree. The angle from her baseline to the tree is  5 35.08. How wide is the river?

tan  < sin  <  5

:

:

61. Two vectors A and B have precisely equal magnitudes. For : : the magnitude of A 1 B to be 100 times larger than the : : magnitude of A 2 B, what must be the angle between them? 62.

:

:

Two vectors A and B have precisely equal magnitudes. : : For the magnitude of A 1 B to be larger than the mag: : nitude of A 2 B by the factor n, what must be the angle between them?

63. There are nearly  3 107 s in one year. Find the percentage error in this approximation, where “percentage error” is defined as

u

Figure P1.55

percentage error 5

u assumed value 2 true value u

3 100% true value 64. An air-traffic controller observes two aircraft on his radar screen. The first is at altitude 800 m, horizontal distance 19.2 km, and 25.08 south of west. The second aircraft is at altitude 1 100 m, horizontal distance 17.6 km, and 20.08 south of west. What is the distance between the two aircraft? (Place the x axis west, the y axis south, and the z axis vertical.) Richard Payne/NASA

Figure P1.56 The Milky Way galaxy.

 1808

where  is in radians and 9 is in degrees. (b) Use a calculator to find the largest angle for which tan  may be approximated by  with an error less than 10.0%.

Additional Problems 56. The distance from the Sun to the nearest star is about 4 3 1016 m. The Milky Way galaxy (Fig. P1.56) is roughly a disk of diameter , 1021 m and thickness , 1019 m.

:

59. Vectors A and B have equal magnitudes of 5.00. The sum of : : A and B is the vector 6.00 ˆj . Determine the angle between : : A and B.

(26.30 b)ˆi 2(4.00 b cos 408)ˆi 2(4.00 b sin 408) ˆj

55.

The consumption of natural gas by a company satisfies the empirical equation V 5 1.50t 1 0.008 00t 2, where V is the volume of gas in millions of cubic feet and t is the time in months. Express this equation in units of cubic feet and seconds. Assume a month is 30.0 days.

65. A child loves to watch as you fill a transparent plastic bottle with shampoo (Fig P1.65, page 34). Every horizontal cross section of the bottle is circular, but the diameters of the circles have different values. You pour the brightly colored shampoo into the bottle at a constant rate of 16.5 cm3/s. At what rate is its level in the bottle rising (a) at a point where the diameter of the bottle is 6.30 cm and (b) at a point where the diameter is 1.35 cm?

34  CHAPTER 1 | Introduction and Vectors What would happen to the answer if you rearranged the order of the trees, for instance, to B (30 m, 220 m), A (60 m, 80 m), E (210 m, 210 m), C (40 m, 230 m), and D (270 m, 60 m)? State reasoning to show that the answer does not depend on the order in which the trees are labeled.

Figure P1.65

66.

67.

68.

69.

One cubic centimeter of water has a mass of 1.00 3 1023 kg. (a) Determine the mass of 1.00 m3 of water. (b) Biological substances are 98% water. Assume that they have the same density as water to estimate the masses of a cell that has a diameter of 1.00 m, a human kidney, and a fly. Model the kidney as a sphere with a radius of 4.00 cm and the fly as a cylinder 4.00 mm long and 2.00 mm in diameter. The helicopter view in Fig. P1.67 shows two people pulling on a stubborn mule. The person on the : right pulls with a force F1 of magnitude 120 N and direction of 1 5 60.08. The person on the left pulls : with a force F2 of magnitude 80.0 N and direction of 2 5 75.08. Find (a) the single force that is equivalent to the two forces shown and (b) the force that a third person would have to exert on the mule to make the resultant force equal to zero. The forces are measured in units of newtons (symbolized N).

S

u

S

u

Figure P1.69

70. You stand in a flat meadow and observe two cows (Fig. P1.70). Cow A is due north of you and 15.0 m from your position. Cow B is 25.0 m from your position. From your point of view, the angle between cow A and cow B is 20.08, with cow B appearing to the right of cow A. (a) How far apart are cow A and cow B? (b) Consider the view seen by cow A. According to this cow, what is the angle between you and cow B? (c) Consider the view seen by cow B. According to this cow, what is the angle between you and cow A? Hint: What does the situation look like to a hummingbird hovering above the meadow? (d) Two stars in the sky appear to be 20.08 apart. Star A is 15.0 ly from the Earth, and star B, appearing to the right of star A, is 25.0 ly from the Earth. To an inhabitant of a planet orbiting star A, what is the angle in the sky between star B and our Sun?

Figure P1.67

A woman wishing to know the height of a mountain measures the angle of elevation of a mountaintop as 12.08. After walking 1.00 km closer to the mountain on level ground, she finds the angle to be 14.08. (a) Draw a picture of the problem, neglecting the height of the woman’s eyes above the ground. Hint: Use two triangles. (b) Using the symbol y to represent the mountain height and the symbol x to represent the woman’s original distance from the mountain, label the picture. (c) Using the labeled picture, write two trigonometric equations relating the two selected variables. (d) Find the height y. A pirate has buried his treasure on an island with five trees located at the points (30.0 m, 220.0 m), (60.0 m, 80.0 m), (210.0 m, 210.0 m), (40.0 m, 230.0 m), and (270.0 m, 60.0 m), all measured relative to some origin, as shown in Figure P1.69. His ship’s log instructs you to start at tree A and move toward tree B, but to cover only one-half the distance between A and B. Then move toward tree C, covering one-third the distance between your current location and C. Next move toward tree D, covering one-fourth the distance between where you are and D. Finally move toward tree E, covering one-fifth the distance between you and E, stop, and dig. (a) Assume you have correctly determined the order in which the pirate labeled the trees as A, B, C, D, and E as shown in the figure. What are the coordinates of the point where his treasure is buried? (b) What If? What if you do not really know the way the pirate labeled the trees?

Figure P1.70 Your view of two cows in a meadow. Cow A is due north of you. You must rotate your eyes through an angle of 20.08 to look from cow A to cow B.

71.

A rectangular parallelepiped has dimensions a, b, and c as shown in Figure P1.71. (a) Obtain a vector expression for : the face diagonal vector R 1 . (b) What is the magnitude of this : : ˆ and R vector? (c) Notice that R 1 , c k, 2 make a right triangle. : Obtain a vector expression for the body diagonal vector R 2.

S

R S

R

Figure P1.71

Context

1

Alternative-Fuel Vehicles T

electric cars with a range of about 20 miles and a top speed of 15 miles per hour had been developed. An internal combustion engine was designed but never built by Dutch physicist Christiaan Huygens in 1680. The invention of modern gasoline-powered internal combustion vehicles is generally credited to Gottlieb Daimler in 1885 and Karl Benz in 1886. Several earlier vehicles, dating back to 1807, however, used internal combustion engines operating on various fuels, including coal gas and primitive gasoline. At the beginning of the 20th century, steam-powered, gasoline-powered, and electric cars shared the roadways in the United States. Electric cars did not possess the vibration, smell, and noise of gasoline-powered cars and did not suffer from the long start-up time intervals, up to 45 minutes, of steam-powered cars on cold mornings. Electric cars were especially preferred by women, who did not enjoy the difficult task of cranking a gasolinepowered car to start the engine. The limited range of electric cars was not a significant problem because the only roads that existed were in highly populated areas and cars were primarily used for short trips in town. The end of electric cars in the early 20th century began with the following developments:

Courtesy of The Exhibition Alliance, Inc., Hamilton, NY

he idea of self-propelled vehicles has been part of the human imagination for centuries. Leonardo da Vinci drew plans for a vehicle powered by a wound spring in 1478. This vehicle was never built although models have been constructed from his plans and appear in museums. Isaac Newton developed a vehicle in 1680 that operated by ejecting steam out the back, similar to a rocket engine. This invention did not develop into a useful device. Despite these and other attempts, self-propelled vehicles did not succeed; that is, they did not begin to replace the horse as a primary means of transportation until the 19th century. The history of successful self-propelled vehicles begins in 1769 with the invention of a military tractor by Nicolas Joseph Cugnot in France. This vehicle, as well as Cugnot’s follow-up vehicles, was powered by a steam engine. During the remainder of the 18th century and for most of the 19th century, additional steam-driven vehicles were developed in France, Great Britain, and the United States. After the invention of the electric battery by Italian Alessandro Volta at the beginning of the 19th century and its further development over three decades came the invention of early electric vehicles in the 1830s. The development in 1859 of the storage battery, which could be recharged, provided significant impetus to the development of electric vehicles. By the early 20th century,

Figure 1 A model of a spring-drive car designed by Leonardo da Vinci.

• 1901: A major discovery of crude oil in Texas reduced prices of gasoline to widely affordable levels. • 1912: The electric starter for gasoline engines was invented, removing the physical task of cranking the engine. • During the 1910s: Henry Ford successfully introduced mass production of internal combustion vehicles, resulting in a drop in the price of these vehicles to significantly less than that of an electric car. • By the early 1920s: Roadways in the United States were of much better quality than in the previous decades, and now connected cities with each other, requiring vehicles with a longer range than when roadways existed only within city limits. Because of these factors, the roadways were ruled by gasoline-powered cars almost exclusively by the 1920s. Gasoline, however, is a finite and short-lived commodity. We are approaching the end of our ability to use gasoline 35

iStockphoto.com/joel-tjoel-t

36  CONTEXT 1 | Alternative-Fuel Vehicles

© Bettmann/CORBIS

Figure 4 Modern electric cars can take advantage of an infrastructure set up in some localities to provide charging stations in parking lots.

Figure 2 This magazine advertisement for an electric car is typical of this popular type of car in the early 20th century.

© Images-USA/Alamy

in transportation; some experts predict that diminishing supplies of crude oil will push the cost of gasoline to prohibitively high levels within two more decades. Furthermore, gasoline and diesel fuel result in serious

tailpipe emissions that are harmful to the environment. As we look for a replacement for gasoline, we also want to pursue fuels that will be kinder to the atmosphere. Such fuels will help reduce the climate change effects of global warming, which we will study in Context 5. What do the steam engine, the electric motor, and the internal combustion engine have in common? That is, what do they each extract from a source, be it a type of fuel or an electric battery? The answer to this question is energy. Regardless of the type of automobile, some source of energy must be provided. Energy is one of the physical concepts that we will investigate in this Context. A fuel such as gasoline contains energy due to its chemical composition and its ability to undergo a combustion process. The battery in an electric car also contains energy, again related to chemical composition, but in this case it is associated with an ability to produce an electric current. One difficult social aspect of developing a new energy source for automobiles is that there must be a synchronized development of the new automobile along with the infrastructure for delivering the new source of energy. This aspect requires close cooperation between automotive corporations and energy manufacturers and suppliers. For example, electric cars cannot be used to travel long distances unless an infrastructure of charging stations develops in parallel with the development of electric cars. As we draw near to the time when we run out of gasoline, our central question in this first Context is an important one for our future development:

Figure 3 A bus running on natural gas operates in Port Huron, Michigan. Several cities such as Port Huron have established natural gas refueling centers so that a large percentage of their fleet can be operated with this fuel that is less expensive than diesel and emits fewer particulates into the atmosphere.

What source besides gasoline can be used to provide energy for an automobile while reducing environmentally damaging emissions?

Chapter

2

Motion in One Dimension Chapter Outline 2.1 Average Velocity 2.2 Instantaneous Velocity 2.3 Analysis Model: Particle Under Constant Velocity 2.4 Acceleration 2.5 Motion Diagrams 2.6 Analysis Model: Particle Under Constant Acceleration 2.7 Freely Falling Objects iStockphoto.com/technotrtechnotr

2.8 Context Connection: Acceleration Required by Consumers SUMMARY

T

o begin our study of motion, it is important to be able to describe motion using the concepts of space and time without regard to the causes of the motion. This portion of mechanics is called kinematics (from the same root as the word cinema). In this chapter, we shall consider motion along a straight line, that is, one-dimensional motion. Chapter 3 extends our discussion to two-dimensional motion. From everyday experience we recognize that motion represents continuous change in the position of an object. For example, if you are driving from your home to a destination, your position on the Earth’s surface is changing. The movement of an object through space (translation) may be accompanied by the rotation or vibration of the object. Such motions can be quite complex. It is often possible to simplify matters, however, by temporarily ignoring rotation and internal motions of the moving object. The result is the simplification model that we call the particle model, discussed in Chapter 1. In many situations, an object can be treated as a particle if the only motion being considered is translation through space. We will use the particle model extensively throughout this book.

One of the physical quantities we will study in this chapter is the velocity of an object moving in a straight line. Downhill skiers can reach velocities with a magnitude greater than 100 km/h.

37

38  CHAPTER 2 | Motion in One Dimension

2.1 | Average Velocity We begin our study of kinematics with the notion of average velocity. You may be familiar with a similar notion, average speed, from experiences with driving. If you drive your car 100 miles according to your odometer and it takes 2.0 hours to do so, your average speed is (100 mi)/(2.0 h) 5 50 mi/h. For a particle moving through a distance d in a time interval Dt, the average speed vavg is mathematically defined as c Definition of average speed

d 2.1b Dt Speed is not a vector, so there is no direction associated with average speed. Average velocity may be a little less familiar to you due to its vector nature. Let us start by imagining the motion of a particle, which, through the particle model, can represent the motion of many types of objects. We shall restrict our study at this point to one-dimensional motion along the x axis. The motion of a particle is completely specified if the position of the particle in space is known at all times. Consider a car moving back and forth along the x axis and imagine that we take data on the position of the car every 10 s. Active Figure 2.1a is a pictorial representation of this one-dimensional motion that shows the positions of the car at 10-s intervals. The six data points we have recorded are represented by the letters 𝖠 through 𝖥. Table 2.1 is a tabular representation of the motion. It lists the data as entries for position at each time. The black dots in Active Figure 2.1b show a graphical representation of the motion. Such a plot is often called a position– time graph. The curved line in Active Figure 2.1b cannot be unambiguously drawn vavg ;

Active Figure 2.1 A car moves back and forth along a straight line. Because we are interested only in the car’s translational motion, we can model it as a particle. Several representations of the information about the motion of the car can be used. Table 2.1 is a tabular representation of the information. (a) A pictorial representation of the motion of the car. (b) A graphical representation, known as a position–time graph, of the car’s motion in part (a). The average velocity vx,avg in the interval t 5 0 to t 5 10 s is obtained from the slope of the straight line connecting points 𝖠 and 𝖡. (c) A velocity–time graph of the motion of the car in part (a).

TABLE 2.1 |

Positions of the Car at Various Times Position

t (s)

x (m)

𝖠

0

30

𝖡

10

52

𝖢

20

38

𝖣

30

0

𝖤

40

237

𝖥

50

253

2.1 | Average Velocity 39

through our six data points because we have no information about what happened between these points. The curved line is, however, a possible graphical representation of the position of the car at all instants of time during the 50 s. If a particle is moving during a time interval Dt 5 tf 2 ti, the displacement of the par: ticle is described as Dx 5: xf 2 : x i 5 (xf 2 xi)ˆi . (Recall from Chapter 1 that displacement is defined as the change in the position of the particle, which is equal to its final position value minus its initial position value.) Because we are considering only onedimensional motion in this chapter, we shall drop the vector notation at this point and pick it up again in Chapter 3. The direction of a vector in this chapter will be indicated by means of a positive or negative sign. The average velocity vx, avg of the particle is defined as the ratio of its displacement Dx to the time interval Dt during which the displacement takes place: vx,avg ;

xf 2 x i Dx 5 Dt tf 2 ti

2.2b

Pitfall Prevention | 2.1

Average Speed and Average Velocity The magnitude of the average velocity is not the average speed. Consider a particle moving from the origin to x 5 10 m and then back to the origin in a time interval of 4.0 s. The magnitude of the average velocity is zero because the particle ends the time interval at the same position at which it started; the displacement is zero. The average speed, however, is the total distance divided by the time interval: 20 m/4.0 s 5 5.0 m/s.

c Definition of average velocity

where the subscript x indicates motion along the x axis. From this definition we see that average velocity has the dimensions of length divided by time: meters per second in SI units and feet per second in U.S. customary units. The average velocity is independent of the path taken between the initial and final points. This independence is a major difference from the average speed discussed at the beginning of this section. The average velocity is independent of path because it is proportional to the displacement Dx, which depends only on the initial and final coordinates of the particle. Average speed (a scalar) is found by dividing the distance traveled by the time interval, whereas average velocity (a vector) is the displacement divided by the time interval. Therefore, average velocity gives us no details of the motion; rather, it only gives us the result of the motion. Finally, note that the average velocity in one dimension can be positive or negative, depending on the sign of the displacement. (The time interval Dt is always positive.) If the x coordinate of the particle increases during the time interval (i.e., if xf . xi), Dx is positive and vx,avg is positive, which corresponds to an average velocity in the positive x direction. On the other hand, if the coordinate decreases over time (xf , xi), Dx is negative; hence, vx,avg is negative, which corresponds to an average velocity in the negative x direction. QUICK QUIZ 2.1 Under which of the following conditions is the magnitude of the average velocity of a particle moving in one dimension smaller than the average speed over some time interval? (a) A particle moves in the 1x direction without reversing. (b) A particle moves in the 2x direction without reversing. (c) A particle moves in the 1x direction and then reverses the direction of its motion. (d) There are no conditions for which it is true.

The average velocity can also be interpreted geometrically, as seen in the graphical representation in Active Figure 2.1b. A straight line can be drawn between any two points on the curve. Active Figure 2.1b shows such a line drawn between points 𝖠 and 𝖡. Using a geometric model, this line forms the hypotenuse of a right triangle of height Dx and base Dt. The slope of the hypotenuse is the ratio Dx/Dt. Therefore, we see that the average velocity of the particle during the time interval ti to tf is equal to the slope of the straight line joining the initial and final points on the position–time graph. For example, the average velocity of the car between points 𝖠 and 𝖡 is vx,avg 5 (52 m 2 30 m)/(10 s 2 0) 5 2.2 m/s. We can also identify a geometric interpretation for the total displacement during the time interval. Active Figure 2.1c shows the velocity–time graphical representation of the motion in Active Figures 2.1a and 2.1b. The total time interval for the motion has been divided into small increments of duration Dtn. During each of these increments, if we model the velocity as constant during the short increment, the displacement of the particle is given by Dxn 5 vn Dtn. Geometrically, the product on the right side of this expression represents the area of a thin rectangle associated with each time increment in Active Figure 2.1c;

Pitfall Prevention | 2.2

Slopes of Graphs In any graph of physical data, the slope represents the ratio of the change in the quantity represented on the vertical axis to the change in the quantity represented on the horizontal axis. Remember that a slope has units (unless both axes have the same units). The units of slope in Active Figures 2.1b and 2.2 (page 42) are meters per second, the units of velocity.

40  CHAPTER 2 | Motion in One Dimension the height of the rectangle (measured from the time axis) is vn, and the width is Dtn. The total displacement of the particle will be the sum of the displacements during each of the increments: Dx < oDxn 5 ovn Dtn n

n

This sum is an approximation because we have modeled the velocity as constant in each increment, which is not the case. The term on the right represents the total area of all the thin rectangles. Now let us take the limit of this expression as the time increments shrink to zero, in which case the approximation becomes exact: Dx 5 lim

oDxn 5 Dtlim ovn Dtn :0 n

Dtn : 0 n

n

In this limit, the sum of the areas of all the very thin rectangles becomes equal to the total area under the curve. Therefore, the displacement of a particle during the time interval ti to tf is equal to the area under the curve between the initial and final points on the velocity–time graph. We will make use of this geometric interpretation in Section 2.6. Example 2.1 | Calculating

the Average Velocity and Speed

Find the displacement, average velocity, and average speed of the car in Active Figure 2.1a between positions 𝖠 and 𝖥.

SOLUTION Conceptualize Consult the pictorial representation in Active Figure 2.1 to form a mental image of the car and its motion. Active Figure 2.1b shows a graphical representation of the motion in the form of a position–time graph for the particle. Categorize We model the car as a particle. We will be substituting numerical values into definitions that we have seen, so this problem will be categorized as a substitution problem. Analyze From the position–time graph given in Active Figure 2.1b, notice that x 𝖠 5 30 m at t 𝖠 5 0 s and that x 𝖥 5 253 m at t𝖥 5 50 s.

Use Equation 1.6 to find the displacement of the car: Use Equation 2.2 to find the car’s average velocity:

Dx 5 x 𝖥 2 x 𝖠 5 253 m 2 30 m 5 283 m vx,avg 5 5

x𝖥 2 x𝖠 t𝖥 2 t𝖠 253 m 2 30 m 283 m 5 5 21.7 m/s 50 s 2 0 s 50 s

We cannot unambiguously find the average speed of the car from the data in Table 2.1, because we do not have information about the positions of the car between the data points. If we adopt the assumption that the details of the car’s position are described by the curve in Active Figure 2.1b, the distance traveled is 22 m (from 𝖠 to 𝖡) plus 105 m (from 𝖡 to 𝖥), for a total of 127 m. Use Equation 2.1 to find the car’s average speed:

vavg 5

127 m 5 2.5 m/s 50 s

Finalize The first result means that the car ends up 83 m in the negative direction (to the left, in this case) from where it started. This number has the correct units and is of the same order of magnitude as the supplied data. A quick look at Active Figure 2.1a indicates that it is the correct answer. The fact that the car ends up to the left of its initial position also makes it reasonable that the average velocity is negative.

Notice that the average speed is positive, as it must be. Suppose the red-brown curve in Active Figure 2.1b were different so that between 0 s and 10 s it went from 𝖠 up to 100 m and then came back down to 𝖡. The average speed of the car would change because the distance is different, but the average velocity would not change. Substitution problems generally do not have an extensive Analyze section other than the substitution of numbers into a given equation. Similarly, the Finalize step consists primarily of checking the units and making sure that the answer is reasonable. Therefore, for substitution problems after this one, we will not label Analyze or Finalize steps. We include those labels in this first example of a substitution problem just to demonstrate the process.

2.2 | Instantaneous Velocity 41

Exampl e 2.2 | Motion

of a Jogger

A jogger runs in a straight line, with an average velocity of magnitude 5.00 m/s for 4.00 min and then with an average velocity of magnitude 4.00 m/s for 3.00 min. (A) What is the magnitude of the final displacement from her initial position?

SOLUTION Conceptualize From your experience, imagine a runner running on a track. Notice that the runner runs more slowly on the average during the second time interval, as if he or she is tiring. Categorize That this problem involves a jogger is not important; we model the jogger as a particle. Analyze From the data for the

two separate portions of the motion, find the displacement for each portion, using Equation 2.2:

vx,avg 5

Dx Dt

: Dx 5 vx,avg Dt

1160mins 2

Dxportion 1 5 (5.00 m/s)(4.00 min) 5 1.20 3 103 m

1160mins 2

Dxportion 2 5 (4.00 m/s)(3.00 min) 5 7.20 3 102 m We add these two displacements to find the total displacement of 1.92 3 103 m. (B) What is the magnitude of her average velocity during this entire time interval of 7.00 min?

SOLUTION Find the average velocity for the entire time interval using Equation 2.2:

vx,avg 5

Dx 1.92 3 103 m 1 min 5 4.57 m/s 5 Dt 7.00 min 60 s

1

2

Finalize Notice that the average velocity is between the two velocities given in the problem, as expected, but is not the

arithmetic mean of these two velocities.

2.2 | Instantaneous Velocity Suppose you drive your car through a displacement of magnitude 40 miles and it takes exactly 1 hour to do so, from 1:00:00 p.m. to 2:00:00 p.m. Then the magnitude of your average velocity is 40 mi/h for the 1-h interval. How fast, though, were you going at the particular instant of time 1:20:00 p.m.? It is likely that your velocity varied during the trip, owing to hills, traffic lights, slow drivers ahead of you, and the like, so that there was not a single velocity maintained during the entire hour of travel. The velocity of a particle at any instant of time is called the instantaneous velocity. Consider again the motion of the car shown in Active Figure 2.1a. Active Figure 2.2a (page 42) is the graphical representation again, with two blue lines representing average velocities over very different time intervals. One blue line represents the average velocity we calculated earlier over the interval from 𝖠 to 𝖡. The second blue line represents the average velocity over the much longer interval 𝖠 to 𝖥. How well does either of these represent the instantaneous velocity at point 𝖠? In Active Figure 2.1a, the car begins to move to the right, which we identify as a positive velocity. The average velocity from 𝖠 to 𝖥 is negative (because the slope of the line from 𝖠 to 𝖥

42  CHAPTER 2 | Motion in One Dimension Active Figure 2.2 (a) Position– time graph for the motion of the car in Active Figure 2.1. (b) An enlargement of the upper left-hand corner of the graph.

Figure 2.3 In the position–time graph shown, the velocity is positive at 𝖠, where the slope of the tangent line is positive; the velocity is zero at 𝖡, where the slope of the tangent line is zero; and the velocity is negative at 𝖢, where the slope of the tangent line is negative.

is negative), so this velocity clearly is not an accurate representation of the instantaneous velocity at 𝖠. The average velocity from interval 𝖠 to 𝖡 is positive, so this velocity at least has the right sign. In Active Figure 2.2b, we show the result of drawing the lines representing the average velocity of the car as point 𝖡 is brought closer and closer to point 𝖠. As that occurs, the slope of the blue line approaches that of the green line, which is the line drawn tangent to the curve at point 𝖠. As 𝖡 approaches 𝖠, the time interval that includes point 𝖠 becomes infinitesimally small. Therefore, the average velocity over this interval as the interval shrinks to zero can be interpreted as the instantaneous velocity at point 𝖠. Furthermore, the slope of the line tangent to the curve at 𝖠 is the instantaneous velocity at the time t 𝖠. In other words, the instantaneous velocity vx equals the limiting value of the ratio Dx/Dt as Dt approaches zero:1 vx ; lim

Dt : 0

Dx Dt

In calculus notation, this limit is called the derivative of x with respect to t, written dx/dt : vx ; lim

c Definition of instantaneous

Dt : 0

velocity

Pitfall Prevention | 2.3

Instantaneous Speed and Instantaneous Velocity In Pitfall Prevention 2.1, we argued that the magnitude of the average velocity is not the average speed. The magnitude of the instantaneous velocity, however, is the instantaneous speed. In an infinitesimal time interval, the magnitude of the displacement is equal to the distance traveled by the particle.

Dx dx 5 Dt dt

2.3b

The instantaneous velocity can be positive, negative, or zero. When the slope of the position–time graph is positive, such as at point 𝖠 in Figure 2.3, vx is positive. At point 𝖢, vx is negative because the slope is negative. Finally, the instantaneous velocity is zero at the peak 𝖡 (the turning point), where the slope is zero. From here on, we shall usually use the word velocity to designate instantaneous velocity. The instantaneous speed of a particle is defined as the magnitude of the instantaneous velocity vector. Hence, by definition, speed can never be negative. QUICK QUIZ 2.2 Are members of the highway patrol more interested in (a) your average speed or (b) your instantaneous speed as you drive?

If you are familiar with calculus, you should recognize that specific rules exist for taking the derivatives of functions. These rules, which are listed in Appendix B.6, enable us to evaluate derivatives quickly. Suppose x is proportional to some power of t, such as x 5 At n where A and n are constants. (This equation is a very common functional form.) The derivative of x with respect to t is dx 5 n At n21 dt For example, if x 5 5t 3, we see that dx/dt 5 3(5)t 321 5 15t 2. 1 Note

that the displacement Dx also approaches zero as Dt approaches zero. As Dx and Dt become smaller and smaller, however, the ratio Dx/Dt approaches a value equal to the true slope of the line tangent to the x versus t curve.

2.2 | Instantaneous Velocity 43

THINKING PHYSICS 2.1

Consider the following motions of an object in one dimension. (a) A ball is thrown directly upward, rises to its highest point, and falls back into the thrower’s hand. (b) A race car starts from rest and speeds up to 100 m/s along a straight line. (c) A spacecraft on the way to another star drifts through empty space at constant velocity. Are there any instants of time in the motion of these objects at which the instantaneous velocity at the instant and the average velocity over the entire interval are the same? If so, identify the point(s). Reasoning (a) The average velocity over the entire interval for the thrown ball

is zero; the ball returns to the starting point at the end of the time interval. There is one point—at the top of the motion—at which the instantaneous velocity is zero. (b) The average velocity for the motion of the race car cannot be evaluated unambiguously with the information given, but its magnitude must be some value between 0 and 100 m/s. Because the magnitude of the instantaneous velocity of the car will have every value between 0 and 100 m/s at some time during the interval, there must be some instant at which the instantaneous velocity is equal to the average velocity over the entire interval. (c) Because the instantaneous velocity of the spacecraft is constant, its instantaneous velocity at any time and its average velocity over any time interval are the same. b

Exampl e 2.3 | The

Limiting Process

The position of a particle moving along the x axis varies in time according to the expression2 x 5 3t 2, where x is in meters and t is in seconds. Find the velocity in terms of t at any time.

SOLUTION Conceptualize The position–time graphical representation for this motion is shown in Figure 2.4. Before beginning the calculation, imagine the motion of the particle on the x axis. Does the particle ever reverse direction? Categorize The entity in motion is already presented as a particle, so no simplification

model is needed. Analyze We can compute the velocity at any time t by using the definition of the instanta-

neous velocity. If the initial coordinate of the particle at time t is xi 5 3t 2, find the coordinate at a later time t 1 Dt: Find the displacement in the time interval Dt: Find the average velocity in this time interval:

xf 5 3(t 1 Dt)2 5 3[t 2 1 2t Dt 1 (Dt)2] 5 3t 2 1 6t Dt 1 3(Dt)2 Dx 5 xf 2 xi 5 (3t 2 1 6t Dt 1 3(Dt)2) 2 (3t 2) 5 6t Dt 1 3(Dt)2 vx,avg 5

6t Dt 1 3(Dt)2 Dx 5 5 6t 1 3 Dt Dt Dt

Dx 5 6t 1 3(0) 5 6t To find the instantaneous velocity, take the vx 5 lim Dt : 0 Dt limit of this expression as Dt approaches zero: continued 2 Simply to make it easier to read, we write the equation as x 5 3t 2 rather than as x 5 (3.00 m/s2)t 2.00. When an equation summarizes measurements, consider its coefficients to have as many significant digits as other data quoted in a problem. Also consider its coefficients to have the units required for dimensional consistency. When we start our clocks at t 5 0, we usually do not mean to limit precision to a single digit. Consider any zero value in this book to have as many significant figures as you need.

Figure 2.4 (Example 2.3) Position–time graph for a particle having an x coordinate that varies in time according to x 5 3t 2. Note that the instantaneous velocity at t 5 3.0 s is obtained from the slope of the green line tangent to the curve at this point.

44  CHAPTER 2 | Motion in One Dimension 2.3 cont. Finalize Notice that this expression gives us the velocity at any general time t. It tells us that vx is increasing linearly in time.

It is then a straightforward matter to find the velocity at some specific time from the expression vx 5 6t by substituting the value of the time. For example, at t 5 3.0 s, the velocity is vx 5 6(3) 5 18 m/s. Again, this answer can be checked from the slope at t 5 3.0 s (the green line in Fig. 2.4). We can also find vx by taking the first derivative of x with respect to time, as in Equation 2.3. In this example, x 5 3t 2, and we see that vx 5 dx/dt 5 6t, in agreement with our result from taking the limit explicitly.

Example 2.4 | Average

and Instantaneous Velocity

A particle moves along the x axis. Its position varies with time according to the expression x 5 24t 1 2t 2, where x is in meters and t is in seconds. The position– time graph for this motion is shown in Figure 2.5a. Because the position of the particle is given by a mathematical function, the motion of the particle is completely known, unlike that of the car in Active Figure 2.1. Notice that the particle moves in the negative x direction for the first second of motion, is momentarily at rest at the moment t 5 1 s, and moves in the positive x direction at times t 1 s. (A) Determine the displacement of the particle in the time intervals t 5 0 to t 5 1 s and t 5 1 s to t 5 3 s.

SOLUTION Conceptualize From the graph in Figure 2.5a, form a mental representation of the particle’s motion. Keep in mind that the particle does not move in a curved path in space such as that shown by the red-brown curve in the graphical representation. The particle moves only along the x axis in one dimension as shown in Figure 2.5b. At t 5 0, is it moving to the right or to the left? During the first time interval, the slope is negative and hence the average velocity is negative. Therefore, we know that the displacement between 𝖠 and 𝖡 must be a negative number having units of meters. Similarly, we expect the displacement between 𝖡 and 𝖣 to be positive. Categorize We will be evaluating results from definitions given in the first two chapters, so we categorize this example as a substitution problem.

Figure 2.5 (Example 2.4) (a) Position–time graph for a particle having an x coordinate that varies in time according to the expression x 5 24t 1 2t 2. (b) The particle moves in one dimension along the x axis.

In the first time interval, set ti 5 t 𝖠 5 0 and tf 5 t 𝖡 5 1 s and use Equation 1.6 to find the displacement:

Dx 𝖠→𝖡 5 xf 2 xi 5 x 𝖡 2 x 𝖠

For the second time interval (t 5 1 s to t 5 3 s), set ti 5 t𝖡 5 1 s and tf 5 t𝖣 5 3 s:

Dx 𝖡 → 𝖣 5 xf 2 xi 5 x 𝖣 2 x 𝖡

5 [24(1) 1 2(1)2] 2 [24(0) 1 2(0)2] 5 22 m 5 [24(3) 1 2(3)2] 2 [24(1) 1 2(1)2] 5 18 m

These displacements can also be read directly from the position–time graph. (B) Calculate the average velocity during these two time intervals.

SOLUTION In the first time interval, use Equation 2.2 with Dt 5 tf 2 ti 5 t 𝖡 2 t 𝖠 5 1 s:

vx,avg(𝖠 → 𝖡) 5

In the second time interval, Dt 5 2 s:

vx,avg(𝖡 → 𝖣) 5

Dx 𝖠 → 𝖡 Dt Dx 𝖡 → 𝖣

Dt These values are the same as the slopes of the blue lines joining these points in Figure 2.5a.

5

22 m 5 22 m/s 1s

5

8m 5 14 m/s 2s

2.3 | Analysis Model: Particle Under Constant Velocity 45

2.4 cont. (C) Find the instantaneous velocity of the particle at t 5 2.5 s.

SOLUTION Measure the slope of the green line at t 5 2.5 s (point 𝖢) in Figure 2.5a:

vx 5

10 m 2 (24 m) 5 16 m/s 3.8 s 2 1.5 s

Notice that this instantaneous velocity is on the same order of magnitude as our previous results, that is, a few meters per second. Is that what you would have expected? Do you see any symmetry in the motion? For example, are there points at which the speed is the same? Is the velocity the same at these points?

2.3 | Analysis Model: Particle Under Constant Velocity As mentioned in Section 1.10, the third category of models used in this book is that of analysis models. Such models help us analyze the situation in a physics problem and guide us toward the solution. An analysis model is a problem we have solved before. It is a description of either (1) the behavior of some physical entity or (2) the interaction between that entity and the environment. When you encounter a new problem, you should identify the fundamental details of the problem and attempt to recognize which, if any, of the types of problems you have already solved might be used as a model for the new problem. This method is somewhat similar to the common practice in the legal profession of finding “legal precedents.” If a previously resolved case can be found that is very similar legally to the present one, it is offered as a model and an argument is made in court to link them logically. The finding in the previous case can then be used to sway the finding in the present case. We will do something similar in physics. For a given problem, we search for a “physics precedent,” a model with which we are already familiar and that can be applied to the present problem. We shall generate analysis models based on four fundamental simplification models. The first simplification model is the particle model discussed in Chapter 1. We will look at a particle under various behaviors and environmental interactions. Further analysis models are introduced in later chapters based on simplification models of a system, a rigid object, and a wave. Once we have introduced these analysis models, we shall see that they appear over and over again later in the book in different situations. When solving a problem, you should avoid browsing through the chapter looking for an equation that contains the unknown variable that is requested in the problem. In many cases, the equation you find may have nothing to do with the problem you are attempting to solve. It is much better to take this first step: Identify the analysis model that is appropriate for the problem. Think carefully about what is going on in the problem and match it to a situation you have seen before. What simplification model is appropriate for the entity involved in the problem? Is it a particle, a system, a rigid object, or a wave? Second, what is the entity doing or how is it interacting with its environment? For example, the analysis model in the title of this section indicates that we modeled the entity of interest as a particle. Furthermore, we determined that the particle is moving with constant velocity. Once the analysis model is identified, there are a small number of equations from which to choose that are appropriate for that model. Therefore, the model tells you which equation(s) to use for the mathematical representation. In this section, we will learn what mathematical equations are associated with the particle under constant

46  CHAPTER 2 | Motion in One Dimension

Figure 2.6 Position–time graph for a particle under constant velocity. The value of the constant velocity is the slope of the line.

velocity analysis model. In the future, when you identify the appropriate model in a problem as the particle under constant velocity, you will immediately know which equations to use to solve the problem. Let us use Equation 2.2 to build our first analysis model. We imagine a particle moving with a constant velocity. The analysis model of a particle under constant velocity can be applied in any situation in which an entity that can be modeled as a particle is moving with constant velocity. This situation occurs frequently, so it is an important model. If the velocity of a particle is constant, its instantaneous velocity at any instant during a time interval is the same as the average velocity over the interval, vx 5 vx,avg. Therefore, we start with Equation 2.2 to generate an equation to be used in the mathematical representation of this situation: vx 5 vx,avg 5

Dx Dt

2.4b

Remembering that Dx 5 xf 2 xi , we see that vx 5 (xf 2 xi)/Dt, or xf 5 xi 1 vx Dt This equation tells us that the position of the particle is given by the sum of its original position xi plus the displacement vx Dt that occurs during the time interval Dt. In practice, we usually choose the time at the beginning of the interval to be ti 5 0 and the time at the end of the interval to be tf 5 t, so our equation becomes c Position as a function of time for the particle under constant velocity model

Example 2.5 | Modeling

xf 5 xi 1 vxt

(for constant vx)

2.5b

Equations 2.4 and 2.5 are the primary equations used in the model of a particle under constant velocity. They can be applied to particles or objects that can be modeled as particles. In the future, once you have identified a problem as requiring the particle under constant velocity model, either of these equations can be used to solve the problem. Figure 2.6 is a graphical representation of the particle under constant velocity. On the position–time graph, the slope of the line representing the motion is constant and equal to the velocity. It is consistent with the mathematical representation, Equation 2.5, which is the equation of a straight line. The slope of the straight line is vx and the y intercept is xi in both representations.

a Runner as a Particle

A kinesiologist is studying the biomechanics of the human body. (Kinesiology is the study of the movement of the human body. Notice the connection to the word kinematics.) She determines the velocity of an experimental subject while he runs along a straight line at a constant rate. The kinesiologist starts the stopwatch at the moment the runner passes a given point and stops it after the runner has passed another point 20 m away. The time interval indicated on the stopwatch is 4.0 s. (A) What is the runner’s velocity?

SOLUTION Conceptualize You have probably watched track and field events at some point in your life, so it should be easy to conceptualize this situation. Categorize We model the moving runner as a particle because the size of the runner and the movement of arms and legs are unnecessary details. Because the problem states that the subject runs at a constant rate, we can model him as a particle under constant velocity. Analyze Having identified the model, we can use Equation 2.4 to find the constant velocity of the runner:

vx 5

xf 2 xi Dx 20 m 2 0 5 5 5 5.0 m/s Dt Dt 4.0 s

2.4 | Acceleration 47

2.5 cont. (B) If the runner continues his motion after the stopwatch is stopped, what is his position after 10 s has passed?

SOLUTION Use Equation 2.5 and the velocity found in part (A) to find the position of the particle at time t 5 10 s:

xf 5 xi 1 vxt 5 0 1 (5.0 m/s)(10 s) 5 50 m

Finalize Is the result for part (A) a reasonable speed for a human? How does it compare to world-record speeds in 100-m and 200-m sprints? Notice that the value in part (B) is more than twice that of the 20-m position at which the stopwatch was stopped. Is this value consistent with the time of 10 s being more than twice the time of 4.0 s?

The mathematical manipulations for the particle under constant velocity stem from Equation 2.4 and its descendent, Equation 2.5. These equations can be used to solve for any variable in the equations that happens to be unknown if the other variables are known. For example, in part (B) of Example 2.5, we find the position when the velocity and the time are known. Similarly, if we know the velocity and the final position, we could use Equation 2.5 to find the time at which the runner is at this position. We shall present more examples of a particle under constant velocity in Chapter 3. A particle under constant velocity moves with a constant speed along a straight line. Now consider a particle moving with a constant speed along a curved path. It can be represented with the particle under constant speed model. The primary equation for this model is Equation 2.1, with the average speed vavg replaced by the constant speed v : v;

d Dt

2.6b

As an example, imagine a particle moving at a constant speed in a circular path. If the speed is 5.00 m/s and the radius of the path is 10.0 m, we can calculate the time interval required to complete one trip around the circle: v5

2␲(10.0 m) d d 2␲r : Dt 5 5 5 12.6 s 5 v v Dt 5.00 m/s

2.4 | Acceleration When the velocity of a particle changes with time, the particle is said to be accelerating. For example, the speed of a car increases when you “step on the gas,” the car slows down when you apply the brakes, and it changes direction when you turn the wheel; these changes are all accelerations. We will need a precise definition of acceleration for our studies of motion. Suppose a particle moving along the x axis has a velocity vxi at time ti and a velocity vxf at time tf. The average acceleration ax, avg of the particle in the time interval Dt 5 tf 2 ti is defined as the ratio Dvx/Dt, where Dvx 5 vxf 2 vxi is the change in velocity of the particle in this time interval: ax,avg ;

vxf 2 vxi tf 2 ti

5

Dvx Dt

2.7b

Therefore, acceleration is a measure of how rapidly the velocity is changing. Acceleration is a vector quantity having dimensions of length divided by (time)2, or L/T2. Some of the common units of acceleration are meters per second per second (m/s2) and feet per second per second (ft/s2). For example, an acceleration of 2 m/s2 means that the velocity changes by 2 m/s during each second of time that passes.

c Definition of average acceleration

48  CHAPTER 2 | Motion in One Dimension In some situations, the value of the average acceleration may be different for different time intervals. It is therefore useful to define the instantaneous acceleration as the limit of the average acceleration as Dt approaches zero, analogous to the definition of instantaneous velocity discussed in Section 2.2: c Definition of instantaneous

ax ; lim

acceleration Pitfall Prevention | 2.4

Negative Acceleration Keep in mind that negative acceleration does not necessarily mean that an object is slowing down. If the acceleration is negative and the velocity is negative, the object is speeding up!

Pitfall Prevention | 2.5

Deceleration The word deceleration has a common popular connotation as slowing down. When combined with the misconception in Pitfall Prevention 2.4 that negative acceleration means slowing down, the situation can be further confused by the use of the word deceleration. We will not use this word in this text.

Dvx

Dt : 0

Dt

5

dvx

2.8b

dt

That is, the instantaneous acceleration equals the derivative of the velocity with respect to time, which by definition is the slope of the velocity–time graph. Note that if ax is positive, the acceleration is in the positive x direction, whereas negative ax implies acceleration in the negative x direction. A negative acceleration does not necessarily mean that the particle is moving in the negative x direction, a point we shall address in more detail shortly. From now on, we use the term acceleration to mean instantaneous acceleration. Because vx 5 dx/dt, the acceleration can also be written ax 5

dvx dt

5

d dx d 2x 5 2 dt dt dt

1 2

2.9b

This equation shows that the acceleration equals the second derivative of the position with respect to time. Figure 2.7 shows how the acceleration–time curve in a graphical representation can be derived from the velocity–time curve. In these diagrams, the acceleration of a particle at any time is simply the slope of the velocity–time graph at that time. Positive values of the acceleration correspond to those points (between 0 and t 𝖡) where the velocity in the positive x direction is increasing in magnitude (the particle is speeding up). The acceleration reaches a maximum at time t 𝖠, when the slope of the velocity–time graph is a maximum. The acceleration then goes to zero at time t 𝖡, when the velocity is a maximum (i.e., when the velocity is momentarily not changing and the slope of the v versus t graph is zero). Finally, the acceleration is negative when the velocity in the positive x direction is decreasing in magnitude (between t 𝖡 and t 𝖢).

QUICK QUIZ 2.3 Using Active Figure 2.8, match each vx–t graph on the top with the ax–t graph on the bottom that best describes the motion.

Active Figure 2.8 (Quick Quiz 2.3) Parts (a), (b), and (c) are velocity–time graphs of objects in one-dimensional motion. The possible acceleration–time graphs of each object are shown in scrambled order in parts (d), (e), and (f).

Figure 2.7 (a) The velocity–time graph for a particle moving along the x axis. (b) The instantaneous acceleration can be obtained from the velocity–time graph.

As an example of the computation of acceleration, consider the pictorial representation of a car’s motion in Figure 2.9. In this case, the velocity of the car has changed from an initial value of 30 m/s to a final value of 15 m/s in a time interval of 2.0 s. The average acceleration during this time interval is ax,avg 5

15 m/s 2 30 m/s 5 27.5 m/s2 2.0 s

2.4 | Acceleration 49

The negative sign in this example indicates that the acceleration vector is in the negative x direction (to the left in Figure 2.9). For the case of motion in a straight line, the direction of the velocity of an object and the direction of its acceleration are related as follows. When the object’s velocity and acceleration are in the same direction, the object is speeding up in that direction. On the other hand, when the object’s velocity and acceleration are in opposite directions, the speed of the object decreases in time. To help with this discussion of the signs of velocity and acceleration, let us take a peek ahead to Chapter 4, where we shall relate the acceleration of an object to the force on the object. We will save the details until that later discussion, but for now, let us borrow the notion that the force on an object is proportional to the acceleration of the object:

S

S

Figure 2.9 The velocity of the car decreases from 30 m/s to 15 m/s in a time interval of 2.0 s.

:

F ~: a

This proportionality indicates that acceleration is caused by force. What’s more, as indicated by the vector notation in the proportionality, force and acceleration are in the same direction. Therefore, let us think about the signs of velocity and acceleration by forming a mental representation in which a force is applied to the object to cause the acceleration. Again consider the case in which the velocity and acceleration are in the same direction. This situation is equivalent to an object moving in a given direction and experiencing a force that pulls on it in the same direction. It is clear in this case that the object speeds up! If the velocity and acceleration are in opposite directions, the object moves one way and a force pulls in the opposite direction. In this case, the object slows down! It is very useful to equate the direction of the acceleration in these situations to the direction of a force because it is easier from our everyday experience to think about what effect a force will have on an object than to think only in terms of the direction of the acceleration. QUICK QUIZ 2.4 If a car is traveling eastward and slowing down, what is the direction of the force on the car that causes it to slow down? (a) eastward (b) westward (c) neither of these directions

Exampl e 2.6 | Average

and Instantaneous Acceleration

The velocity of a particle moving along the x axis varies according to the expression vx 5 40 2 5t 2, where vx is in meters per second and t is in seconds. (A) Find the average acceleration in the time interval t 5 0 to t 5 2.0 s.

SOLUTION Conceptualize Think about what the particle is doing from the mathematical representation. Is it moving at t 5 0? In which direction? Does it speed up or slow down? Figure 2.10 is a vx2t graph that was created from the velocity versus time expression given in the problem statement. Because the slope of the entire vx2t curve is negative, we expect the acceleration to be negative.

Figure 2.10 (Example 2.6) The velocity–time graph for a particle moving along the x axis according to the expression vx 5 40 2 5t 2.

Categorize While this problem does not involve an analysis model, it does involve taking a limit of a function, so it is a bit more sophisticated than a pure substitution problem. Analyze Find the velocities at ti 5 t 𝖠 5 0 and tf 5 t 𝖡 5 2.0 s by

substituting these values of t into the expression for the velocity:

vx 𝖠 5 40 2 5t 𝖠2 5 40 2 5(0)2 5 140 m/s vx 𝖡 5 40 2 5t 𝖡2 5 40 2 5(2.0)2 5 120 m/s continued

50  CHAPTER 2 | Motion in One Dimension 2.6 cont. Find the average acceleration in the specified time interval Dt 5 t 𝖡 2 t 𝖠 5 2.0 s:

ax,avg 5

vxf 2 vxi tf 2 ti

5

vx 𝖡 2 vx 𝖠 t𝖡 2 t𝖠

5

20 m/s 2 40 m/s 2.0 s 2 0 s

5 210 m/s2 Finalize The negative sign is consistent with our expectations: the average acceleration, represented by the slope of the blue

line joining the initial and final points on the velocity–time graph, is negative. (B) Determine the acceleration at t 5 2.0 s.

SOLUTION Analyze

Knowing that the initial velocity at any time t is vxi 5 40 2 5t 2, find the velocity at any later time t 1 Dt: Find the change in velocity over the time interval Dt:

vxf 5 40 2 5(t 1 Dt)2 5 40 2 5t 2 2 10t Dt 2 5(Dt)2 Dvx 5 vxf 2 vxi 5 210t Dt 2 5(Dt)2 Dvx

To find the acceleration at any time t, divide this expression by Dt and take the limit of the result as Dt approaches zero:

ax 5 lim

Substitute t 5 2.0 s:

ax 5 (210)(2.0) m/s2 5 220 m/s2

Dt : 0

Dt

5 lim (210t 2 5Dt) 5 210t Dt : 0

Finalize Because the velocity of the particle is positive and the acceleration is negative at this instant, the particle is slowing down.

Notice that the answers to parts (A) and (B) are different. The average acceleration in part (A) is the slope of the blue line in Figure 2.10 connecting points 𝖠 and 𝖡. The instantaneous acceleration in part (B) is the slope of the green line tangent to the curve at point 𝖡. Notice also that the acceleration is not constant in this example. Situations involving constant acceleration are treated in Section 2.6.

2.5 | Motion Diagrams The concepts of velocity and acceleration are often confused with each other, but in fact they are quite different quantities. It is instructive to make use of the specialized pictorial representation called a motion diagram to describe the velocity and acceleration vectors while an object is in motion. A stroboscopic photograph of a moving object shows several images of the object taken as the strobe light flashes at a constant rate. Figure 2.1a is a motion diagram for the car studied in Section 2.1. Active Figure 2.11 represents three sets of strobe photographs of cars moving along a straight roadway in a single direction, from left to right. The time intervals between flashes of the stroboscope are equal in each part of the diagram. To distinguish between the two vector quantities, we use red arrows for velocity vectors and purple arrows for acceleration vectors in Active Figure 2.11. The vectors are sketched at several instants during the motion of the object. Let us describe the motion of the car in each diagram. In Active Figure 2.11a, the images of the car are equally spaced, and the car moves through the same displacement in each time interval. Therefore, the car moves with constant positive velocity and has zero acceleration. We could model the car as a particle and describe it using the particle under constant velocity analysis model. In Active Figure 2.11b, the images of the car become farther apart as time progresses. In this case, the velocity vector increases in time because the car’s displacement between adjacent positions increases as time progresses. Therefore, the car is moving with a positive velocity and a positive acceleration. The velocity and acceleration are in the same direction. In terms of our earlier force discussion, imagine a force pulling on the car in the same direction it is moving: it speeds up. In Active Figure 2.11c, we interpret the car as slowing down as it moves to the right because its displacement between adjacent positions decreases as time progresses. In

2.6 | Analysis Model: Particle Under Constant Acceleration 51

Active Figure 2.11 Motion diagrams of a car moving along a straight roadway in a single direction. The velocity at each instant is indicated by a red arrow, and the constant acceleration is indicated by a purple arrow.

this case, the car moves initially to the right with a positive velocity and a negative acceleration. The velocity vector decreases in time and eventually reaches zero. (This type of motion is exhibited by a car that skids to a stop after its brakes are applied.) From this diagram we see that the acceleration and velocity vectors are not in the same direction. The velocity and acceleration are in opposite directions. In terms of our earlier force discussion, imagine a force pulling on the car opposite to the direction it is moving: it slows down. The purple acceleration vectors in Active Figures 2.11b and 2.11c are all the same length. Therefore, these diagrams represent a motion with constant acceleration. This important type of motion is discussed in the next section. QUICK QUIZ 2.5 Which of the following statements is true? (a) If a car is traveling eastward, its acceleration must be eastward. (b) If a car is slowing down, its acceleration must be negative. (c) A particle with constant acceleration can never stop and stay stopped.

2.6 | Analysis Model: Particle Under Constant Acceleration If the acceleration of a particle varies in time, the motion may be complex and difficult to analyze. A very common and simple type of one-dimensional motion occurs when the acceleration is constant, such as for the motion of the cars in Active Figures 2.11b and 2.11c. In this case, the average acceleration over any time interval equals the instantaneous acceleration at any instant of time within the interval. Consequently, the velocity increases or decreases at the same rate throughout the motion. The particle under constant acceleration is a common analysis model that we can apply to appropriate problems. It is often used to model situations such as falling objects and braking cars. If we replace ax,avg with the constant ax in Equation 2.7, we find that ax 5

vxf 2 vxi

Active Figure 2.12 Graphical representations of a particle moving along the x axis with constant acceleration ax. (a) The position–time graph, (b) the velocity–time graph, and (c) the acceleration–time graph.

tf 2 ti For convenience, let ti 5 0 and tf be any arbitrary time t. With this notation, we can solve for vxf : vxf 5 vxi 1 axt

(for constant ax)

2.10b

This expression enables us to predict the velocity at any time t if the initial velocity and constant acceleration are known. It is the first of four equations that can be used to solve problems using the particle under constant acceleration model. A graphical representation of position versus time for this motion is shown in Active Figure 2.12a. The velocity–time graph shown in Active Figure 2.12b is a straight line, the slope of

c Velocity as a function of time for the particle under constant acceleration model

52  CHAPTER 2 | Motion in One Dimension which is the constant acceleration ax. The straight line on this graph is consistent with ax 5 dvx /dt being a constant. From this graph and from Equation 2.10, we see that the velocity at any time t is the sum of the initial velocity vxi and the change in velocity axt due to the acceleration. The graph of acceleration versus time (Active Fig. 2.12c) is a straight line with a slope of zero because the acceleration is constant. If the acceleration were negative, the slope of Active Figure 2.12b would be negative and the horizontal line in Active Figure 2.12c would be below the time axis. We can generate another equation for the particle under constant acceleration model by recalling a result from Section 2.1 that the displacement of a particle is the area under the curve on a velocity – time graph. Because the velocity varies linearly with time (see Active Fig. 2.12b), the area under the curve is the sum of a rectangular area (under the horizontal dashed line in Active Fig. 2.12b) and a triangular area (from the horizontal dashed line upward to the curve). Therefore, Dx 5 vxi Dt 1 12(vxf 2 vxi )Dt which can be simplified as follows: Dx 5 (vxi 1 12vxf 2 12vxi)Dt 5 12(vxi 1 vxf )Dt In general, from Equation 2.2, the displacement for a time interval is Dx 5 vx, avgDt Comparing these last two equations, we find that the average velocity in any time interval is the arithmetic mean of the initial velocity vxi and the final velocity vxf : c Average velocity for the particle under constant acceleration model

vx, avg 5 12(vxi 1 vxf )

(for constant ax)

2.11b

Remember that this expression is valid only when the acceleration is constant, that is, when the velocity varies linearly with time. We now use Equations 2.2 and 2.11 to obtain the position as a function of time. Again we choose ti 5 0, at which time the initial position is x i , which gives Dx 5 vx, avg Dt 5 12(vxi 1 vxf )t xf 5 xi 1 12(vxi 1 vxf )t

c Position as a function of velocity and time for the particle under constant acceleration model

(for constant ax)

2.12b

We can obtain another useful expression for the position by substituting Equation 2.10 for vxf in Equation 2.12: xf 5 xi 1 12[vxi 1 (vxi 1 axt)]t xf 5 xi 1 vxit 1 12axt 2

c Position as a function of time for the particle under constant acceleration model

for the particle under constant acceleration model

2.13b

Note that the position at any time t is the sum of the initial position x i , the displacement vxi t that would result if the velocity remained constant at the initial velocity, and the displacement 12axt 2 because the particle is accelerating. Consider again the position – time graph for motion under constant acceleration shown in Active Figure 2.12a. The curve representing Equation 2.13 is a parabola, as shown by the t 2 dependence in the equation. The slope of the tangent to this curve at t 5 0 equals the initial velocity vxi , and the slope of the tangent line at any time t equals the velocity at that time. Finally, we can obtain an expression that does not contain the time by substituting the value of t from Equation 2.10 into Equation 2.12, which gives xf 5 xi 1

c Velocity as a function of position

(for constant ax)

1 2 (vxi

1 vxf)

1

vxf 2 vxi ax

vxf 2 5 vxi 2 1 2ax(xf 2 xi)

2 5x 1 i

vxf 2 2 vxi 2 2ax

(for constant ax)

2.14b

2.6 | Analysis Model: Particle Under Constant Acceleration 53 TABLE 2.2 | Kinematic Equations for Motion of a Particle Under Constant

Acceleration Equation Number

Equation

2.10

vxf 5 vxi 1 axt

2.12

xf 5 xi 1

Information Given by Equation 1 2 (vxf

Velocity as a function of time 1 vxi)t

Position as a function of velocity and time

1 2 2 axt

2.13

xf 5 xi 1 vxit 1

2.14

vxf 2 5 vxi 2 1 2ax(xf 2 xi)

Position as a function of time Velocity as a function of position

Note: Motion is along the x axis. At t 5 0, the position of the particle is xi and its velocity is vxi .

This expression is not an independent equation because it arises from combining Equations 2.10 and 2.12. It is useful, however, for those problems in which a value for the time is not involved. If motion occurs in which the constant value of the acceleration is zero, Equations 2.10 and 2.13 become vxf 5 vxi xf 5 xi 1 vxit

6

when ax 5 0

That is, when the acceleration is zero, the velocity remains constant and the position changes linearly with time. In this case, the particle under constant acceleration model reduces to the particle under constant velocity model. Equations 2.10, 2.12, 2.13, and 2.14 are four kinematic equations that may be used to solve any problem in one-dimensional motion of a particle (or an object that can be modeled as a particle) under constant acceleration. If your analysis of a problem indicates that the particle under constant acceleration is the appropriate analysis model, select from these four equations to solve the problem. Keep in mind that these relationships were derived from the definitions of velocity and acceleration together with some simple algebraic manipulations and the requirement that the acceleration be constant. It is often convenient to choose the initial position of the particle as the origin of the motion so that xi 5 0 at t 5 0. We will see cases, however, in which we must choose the value of xi to be something other than zero. The four kinematic equations for the particle under constant acceleration are listed in Table 2.2 for convenience. The choice of which kinematic equation or equations you should use in a given situation depends on what is known beforehand. Sometimes it is necessary to use two of these equations to solve for two unknowns, such as the position and velocity at some instant. You should recognize that the quantities that vary during the motion are velocity vxf , position xf , and time t. The other quantities — x i , vxi , and ax — are parameters of the motion and remain constant.

PROBLEM-SOLVING STRATEGY: Particle Under Constant Acceleration

The following procedure is recommended for solving problems that involve an object undergoing a constant acceleration. As mentioned in Chapter 1, individual strategies such as this one will follow the outline of the General ProblemSolving Strategy from Chapter 1, with specific hints regarding the application of the general strategy to the material in the individual chapters. 1. Conceptualize Think about what is going on physically in the problem.

Establish the mental representation. 2. Categorize Simplify the problem as much as possible. Confirm that the problem involves either a particle or an object that can be modeled as a particle and that it is moving with a constant acceleration. Construct an appropriate

54  CHAPTER 2 | Motion in One Dimension pictorial representation, such as a motion diagram, or a graphical representation. Make sure all the units in the problem are consistent. That is, if positions are measured in meters, be sure that velocities have units of m/s and accelerations have units of m/s2. Choose a coordinate system to be used throughout the problem. 3. Analyze Set up the mathematical representation. Choose an instant to call the “initial” time t 5 0 and another to call the “final” time t. Let your choice be guided by what you know about the particle and what you want to know about it. The initial instant need not be when the particle starts to move, and the final instant will only rarely be when the particle stops moving. Identify all the quantities given in the problem and a separate list of those to be determined. A tabular representation of these quantities may be helpful to you. Select from the list of kinematic equations the one or ones that will enable you to determine the unknowns. Solve the equations. 4. Finalize Once you have determined your result, check to see if your answers are consistent with the mental and pictorial representations and that your results are realistic.

Example 2.7 | Carrier

Landing

A jet lands on an aircraft carrier at a speed of 140 mi/h (< 63 m/s). (A) What is its acceleration (assumed constant) if it stops in 2.0 s due to an arresting cable that snags the jet and brings it to a stop?

SOLUTION Conceptualize You might have seen movies or television shows in which a jet lands on an aircraft carrier and is brought to rest surprisingly fast by an arresting cable. A careful reading of the problem reveals that in addition to being given the initial speed of 63 m/s, we also know that the final speed is zero. We define our x axis as the direction of motion of the jet. Notice that we have no information about the change in position of the jet while it is slowing down. Categorize Because the acceleration of the jet is assumed constant, we model it as a particle under constant acceleration. Analyze

Equation 2.10 is the only equation in Table 2.2 that does not involve position, so we use it to find the acceleration of the jet, modeled as a particle:

ax 5

vxf 2 vxi t