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  1. Preface
  2. 1 Introduction: The Nature of Science and Physics
    1. Connection for AP® Courses
    2. 1.1 Physics: An Introduction
    3. 1.2 Physical Quantities and Units
    4. 1.3 Accuracy, Precision, and Significant Figures
    5. 1.4 Approximation
    6. Glossary
    7. Section Summary
    8. Conceptual Questions
    9. Problems & Exercises
  3. 2 Kinematics
    1. Connection for AP® Courses
    2. 2.1 Displacement
    3. 2.2 Vectors, Scalars, and Coordinate Systems
    4. 2.3 Time, Velocity, and Speed
    5. 2.4 Acceleration
    6. 2.5 Motion Equations for Constant Acceleration in One Dimension
    7. 2.6 Problem-Solving Basics for One Dimensional Kinematics
    8. 2.7 Falling Objects
    9. 2.8 Graphical Analysis of One Dimensional Motion
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  4. 3 Two-Dimensional Kinematics
    1. Connection for AP® Courses
    2. 3.1 Kinematics in Two Dimensions: An Introduction
    3. 3.2 Vector Addition and Subtraction: Graphical Methods
    4. 3.3 Vector Addition and Subtraction: Analytical Methods
    5. 3.4 Projectile Motion
    6. 3.5 Addition of Velocities
    7. Glossary
    8. Section Summary
    9. Conceptual Questions
    10. Problems & Exercises
    11. Test Prep for AP® Courses
  5. 4 Dynamics: Force and Newton's Laws of Motion
    1. Connection for AP® Courses
    2. 4.1 Development of Force Concept
    3. 4.2 Newton's First Law of Motion: Inertia
    4. 4.3 Newton's Second Law of Motion: Concept of a System
    5. 4.4 Newton's Third Law of Motion: Symmetry in Forces
    6. 4.5 Normal, Tension, and Other Examples of Force
    7. 4.6 Problem-Solving Strategies
    8. 4.7 Further Applications of Newton's Laws of Motion
    9. 4.8 Extended Topic: The Four Basic Forces—An Introduction
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  6. 5 Further Applications of Newton's Laws: Friction, Drag, and Elasticity
    1. Connection for AP® Courses
    2. 5.1 Friction
    3. 5.2 Drag Forces
    4. 5.3 Elasticity: Stress and Strain
    5. Glossary
    6. Section Summary
    7. Conceptual Questions
    8. Problems & Exercises
    9. Test Prep for AP® Courses
  7. 6 Gravitation and Uniform Circular Motion
    1. Connection for AP® Courses
    2. 6.1 Rotation Angle and Angular Velocity
    3. 6.2 Centripetal Acceleration
    4. 6.3 Centripetal Force
    5. 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
    6. 6.5 Newton's Universal Law of Gravitation
    7. 6.6 Satellites and Kepler's Laws: An Argument for Simplicity
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  8. 7 Work, Energy, and Energy Resources
    1. Connection for AP® Courses
    2. 7.1 Work: The Scientific Definition
    3. 7.2 Kinetic Energy and the Work-Energy Theorem
    4. 7.3 Gravitational Potential Energy
    5. 7.4 Conservative Forces and Potential Energy
    6. 7.5 Nonconservative Forces
    7. 7.6 Conservation of Energy
    8. 7.7 Power
    9. 7.8 Work, Energy, and Power in Humans
    10. 7.9 World Energy Use
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  9. 8 Linear Momentum and Collisions
    1. Connection for AP® courses
    2. 8.1 Linear Momentum and Force
    3. 8.2 Impulse
    4. 8.3 Conservation of Momentum
    5. 8.4 Elastic Collisions in One Dimension
    6. 8.5 Inelastic Collisions in One Dimension
    7. 8.6 Collisions of Point Masses in Two Dimensions
    8. 8.7 Introduction to Rocket Propulsion
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  10. 9 Statics and Torque
    1. Connection for AP® Courses
    2. 9.1 The First Condition for Equilibrium
    3. 9.2 The Second Condition for Equilibrium
    4. 9.3 Stability
    5. 9.4 Applications of Statics, Including Problem-Solving Strategies
    6. 9.5 Simple Machines
    7. 9.6 Forces and Torques in Muscles and Joints
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  11. 10 Rotational Motion and Angular Momentum
    1. Connection for AP® Courses
    2. 10.1 Angular Acceleration
    3. 10.2 Kinematics of Rotational Motion
    4. 10.3 Dynamics of Rotational Motion: Rotational Inertia
    5. 10.4 Rotational Kinetic Energy: Work and Energy Revisited
    6. 10.5 Angular Momentum and Its Conservation
    7. 10.6 Collisions of Extended Bodies in Two Dimensions
    8. 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  12. 11 Fluid Statics
    1. Connection for AP® Courses
    2. 11.1 What Is a Fluid?
    3. 11.2 Density
    4. 11.3 Pressure
    5. 11.4 Variation of Pressure with Depth in a Fluid
    6. 11.5 Pascal’s Principle
    7. 11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
    8. 11.7 Archimedes’ Principle
    9. 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
    10. 11.9 Pressures in the Body
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  13. 12 Fluid Dynamics and Its Biological and Medical Applications
    1. Connection for AP® Courses
    2. 12.1 Flow Rate and Its Relation to Velocity
    3. 12.2 Bernoulli’s Equation
    4. 12.3 The Most General Applications of Bernoulli’s Equation
    5. 12.4 Viscosity and Laminar Flow; Poiseuille’s Law
    6. 12.5 The Onset of Turbulence
    7. 12.6 Motion of an Object in a Viscous Fluid
    8. 12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  14. 13 Temperature, Kinetic Theory, and the Gas Laws
    1. Connection for AP® Courses
    2. 13.1 Temperature
    3. 13.2 Thermal Expansion of Solids and Liquids
    4. 13.3 The Ideal Gas Law
    5. 13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
    6. 13.5 Phase Changes
    7. 13.6 Humidity, Evaporation, and Boiling
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  15. 14 Heat and Heat Transfer Methods
    1. Connection for AP® Courses
    2. 14.1 Heat
    3. 14.2 Temperature Change and Heat Capacity
    4. 14.3 Phase Change and Latent Heat
    5. 14.4 Heat Transfer Methods
    6. 14.5 Conduction
    7. 14.6 Convection
    8. 14.7 Radiation
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  16. 15 Thermodynamics
    1. Connection for AP® Courses
    2. 15.1 The First Law of Thermodynamics
    3. 15.2 The First Law of Thermodynamics and Some Simple Processes
    4. 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
    5. 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
    6. 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
    7. 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
    8. 15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  17. 16 Oscillatory Motion and Waves
    1. Connection for AP® Courses
    2. 16.1 Hooke’s Law: Stress and Strain Revisited
    3. 16.2 Period and Frequency in Oscillations
    4. 16.3 Simple Harmonic Motion: A Special Periodic Motion
    5. 16.4 The Simple Pendulum
    6. 16.5 Energy and the Simple Harmonic Oscillator
    7. 16.6 Uniform Circular Motion and Simple Harmonic Motion
    8. 16.7 Damped Harmonic Motion
    9. 16.8 Forced Oscillations and Resonance
    10. 16.9 Waves
    11. 16.10 Superposition and Interference
    12. 16.11 Energy in Waves: Intensity
    13. Glossary
    14. Section Summary
    15. Conceptual Questions
    16. Problems & Exercises
    17. Test Prep for AP® Courses
  18. 17 Physics of Hearing
    1. Connection for AP® Courses
    2. 17.1 Sound
    3. 17.2 Speed of Sound, Frequency, and Wavelength
    4. 17.3 Sound Intensity and Sound Level
    5. 17.4 Doppler Effect and Sonic Booms
    6. 17.5 Sound Interference and Resonance: Standing Waves in Air Columns
    7. 17.6 Hearing
    8. 17.7 Ultrasound
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  19. 18 Electric Charge and Electric Field
    1. Connection for AP® Courses
    2. 18.1 Static Electricity and Charge: Conservation of Charge
    3. 18.2 Conductors and Insulators
    4. 18.3 Conductors and Electric Fields in Static Equilibrium
    5. 18.4 Coulomb’s Law
    6. 18.5 Electric Field: Concept of a Field Revisited
    7. 18.6 Electric Field Lines: Multiple Charges
    8. 18.7 Electric Forces in Biology
    9. 18.8 Applications of Electrostatics
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  20. 19 Electric Potential and Electric Field
    1. Connection for AP® Courses
    2. 19.1 Electric Potential Energy: Potential Difference
    3. 19.2 Electric Potential in a Uniform Electric Field
    4. 19.3 Electrical Potential Due to a Point Charge
    5. 19.4 Equipotential Lines
    6. 19.5 Capacitors and Dielectrics
    7. 19.6 Capacitors in Series and Parallel
    8. 19.7 Energy Stored in Capacitors
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  21. 20 Electric Current, Resistance, and Ohm's Law
    1. Connection for AP® Courses
    2. 20.1 Current
    3. 20.2 Ohm’s Law: Resistance and Simple Circuits
    4. 20.3 Resistance and Resistivity
    5. 20.4 Electric Power and Energy
    6. 20.5 Alternating Current versus Direct Current
    7. 20.6 Electric Hazards and the Human Body
    8. 20.7 Nerve Conduction–Electrocardiograms
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  22. 21 Circuits, Bioelectricity, and DC Instruments
    1. Connection for AP® Courses
    2. 21.1 Resistors in Series and Parallel
    3. 21.2 Electromotive Force: Terminal Voltage
    4. 21.3 Kirchhoff’s Rules
    5. 21.4 DC Voltmeters and Ammeters
    6. 21.5 Null Measurements
    7. 21.6 DC Circuits Containing Resistors and Capacitors
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  23. 22 Magnetism
    1. Connection for AP® Courses
    2. 22.1 Magnets
    3. 22.2 Ferromagnets and Electromagnets
    4. 22.3 Magnetic Fields and Magnetic Field Lines
    5. 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
    6. 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
    7. 22.6 The Hall Effect
    8. 22.7 Magnetic Force on a Current-Carrying Conductor
    9. 22.8 Torque on a Current Loop: Motors and Meters
    10. 22.9 Magnetic Fields Produced by Currents: Ampere’s Law
    11. 22.10 Magnetic Force between Two Parallel Conductors
    12. 22.11 More Applications of Magnetism
    13. Glossary
    14. Section Summary
    15. Conceptual Questions
    16. Problems & Exercises
    17. Test Prep for AP® Courses
  24. 23 Electromagnetic Induction, AC Circuits, and Electrical Technologies
    1. Connection for AP® Courses
    2. 23.1 Induced Emf and Magnetic Flux
    3. 23.2 Faraday’s Law of Induction: Lenz’s Law
    4. 23.3 Motional Emf
    5. 23.4 Eddy Currents and Magnetic Damping
    6. 23.5 Electric Generators
    7. 23.6 Back Emf
    8. 23.7 Transformers
    9. 23.8 Electrical Safety: Systems and Devices
    10. 23.9 Inductance
    11. 23.10 RL Circuits
    12. 23.11 Reactance, Inductive and Capacitive
    13. 23.12 RLC Series AC Circuits
    14. Glossary
    15. Section Summary
    16. Conceptual Questions
    17. Problems & Exercises
    18. Test Prep for AP® Courses
  25. 24 Electromagnetic Waves
    1. Connection for AP® Courses
    2. 24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
    3. 24.2 Production of Electromagnetic Waves
    4. 24.3 The Electromagnetic Spectrum
    5. 24.4 Energy in Electromagnetic Waves
    6. Glossary
    7. Section Summary
    8. Conceptual Questions
    9. Problems & Exercises
    10. Test Prep for AP® Courses
  26. 25 Geometric Optics
    1. Connection for AP® Courses
    2. 25.1 The Ray Aspect of Light
    3. 25.2 The Law of Reflection
    4. 25.3 The Law of Refraction
    5. 25.4 Total Internal Reflection
    6. 25.5 Dispersion: The Rainbow and Prisms
    7. 25.6 Image Formation by Lenses
    8. 25.7 Image Formation by Mirrors
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  27. 26 Vision and Optical Instruments
    1. Connection for AP® Courses
    2. 26.1 Physics of the Eye
    3. 26.2 Vision Correction
    4. 26.3 Color and Color Vision
    5. 26.4 Microscopes
    6. 26.5 Telescopes
    7. 26.6 Aberrations
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  28. 27 Wave Optics
    1. Connection for AP® Courses
    2. 27.1 The Wave Aspect of Light: Interference
    3. 27.2 Huygens's Principle: Diffraction
    4. 27.3 Young’s Double Slit Experiment
    5. 27.4 Multiple Slit Diffraction
    6. 27.5 Single Slit Diffraction
    7. 27.6 Limits of Resolution: The Rayleigh Criterion
    8. 27.7 Thin Film Interference
    9. 27.8 Polarization
    10. 27.9 *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  29. 28 Special Relativity
    1. Connection for AP® Courses
    2. 28.1 Einstein’s Postulates
    3. 28.2 Simultaneity And Time Dilation
    4. 28.3 Length Contraction
    5. 28.4 Relativistic Addition of Velocities
    6. 28.5 Relativistic Momentum
    7. 28.6 Relativistic Energy
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  30. 29 Introduction to Quantum Physics
    1. Connection for AP® Courses
    2. 29.1 Quantization of Energy
    3. 29.2 The Photoelectric Effect
    4. 29.3 Photon Energies and the Electromagnetic Spectrum
    5. 29.4 Photon Momentum
    6. 29.5 The Particle-Wave Duality
    7. 29.6 The Wave Nature of Matter
    8. 29.7 Probability: The Heisenberg Uncertainty Principle
    9. 29.8 The Particle-Wave Duality Reviewed
    10. Glossary
    11. Section Summary
    12. Conceptual Questions
    13. Problems & Exercises
    14. Test Prep for AP® Courses
  31. 30 Atomic Physics
    1. Connection for AP® Courses
    2. 30.1 Discovery of the Atom
    3. 30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
    4. 30.3 Bohr’s Theory of the Hydrogen Atom
    5. 30.4 X Rays: Atomic Origins and Applications
    6. 30.5 Applications of Atomic Excitations and De-Excitations
    7. 30.6 The Wave Nature of Matter Causes Quantization
    8. 30.7 Patterns in Spectra Reveal More Quantization
    9. 30.8 Quantum Numbers and Rules
    10. 30.9 The Pauli Exclusion Principle
    11. Glossary
    12. Section Summary
    13. Conceptual Questions
    14. Problems & Exercises
    15. Test Prep for AP® Courses
  32. 31 Radioactivity and Nuclear Physics
    1. Connection for AP® Courses
    2. 31.1 Nuclear Radioactivity
    3. 31.2 Radiation Detection and Detectors
    4. 31.3 Substructure of the Nucleus
    5. 31.4 Nuclear Decay and Conservation Laws
    6. 31.5 Half-Life and Activity
    7. 31.6 Binding Energy
    8. 31.7 Tunneling
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  33. 32 Medical Applications of Nuclear Physics
    1. Connection for AP® Courses
    2. 32.1 Medical Imaging and Diagnostics
    3. 32.2 Biological Effects of Ionizing Radiation
    4. 32.3 Therapeutic Uses of Ionizing Radiation
    5. 32.4 Food Irradiation
    6. 32.5 Fusion
    7. 32.6 Fission
    8. 32.7 Nuclear Weapons
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
    13. Test Prep for AP® Courses
  34. 33 Particle Physics
    1. Connection for AP® Courses
    2. 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
    3. 33.2 The Four Basic Forces
    4. 33.3 Accelerators Create Matter from Energy
    5. 33.4 Particles, Patterns, and Conservation Laws
    6. 33.5 Quarks: Is That All There Is?
    7. 33.6 GUTs: The Unification of Forces
    8. Glossary
    9. Section Summary
    10. Conceptual Questions
    11. Problems & Exercises
    12. Test Prep for AP® Courses
  35. 34 Frontiers of Physics
    1. Connection for AP® Courses
    2. 34.1 Cosmology and Particle Physics
    3. 34.2 General Relativity and Quantum Gravity
    4. 34.3 Superstrings
    5. 34.4 Dark Matter and Closure
    6. 34.5 Complexity and Chaos
    7. 34.6 High-Temperature Superconductors
    8. 34.7 Some Questions We Know to Ask
    9. Glossary
    10. Section Summary
    11. Conceptual Questions
    12. Problems & Exercises
  36. A | Atomic Masses
  37. B | Selected Radioactive Isotopes
  38. C | Useful Information
  39. D | Glossary of Key Symbols and Notation
  40. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
    18. Chapter 18
    19. Chapter 19
    20. Chapter 20
    21. Chapter 21
    22. Chapter 22
    23. Chapter 23
    24. Chapter 24
    25. Chapter 25
    26. Chapter 26
    27. Chapter 27
    28. Chapter 28
    29. Chapter 29
    30. Chapter 30
    31. Chapter 31
    32. Chapter 32
    33. Chapter 33
    34. Chapter 34
  41. Index
An airplane flying very low to the ground, just above a beach full of onlookers, as it comes in for a landing.
Figure 2.12 A plane decelerates, or slows down, as it comes in for landing in St. Maarten. Its acceleration is opposite in direction to its velocity. (credit: Steve Conry, Flickr)

Learning Objectives

By the end of this section, you will be able to:

  • Define and distinguish between instantaneous acceleration and average acceleration.
  • Calculate acceleration given initial time, initial velocity, final time, and final velocity.

The information presented in this section supports the following AP® learning objectives and science practices:

  • 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2)
  • 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical representations. (S.P. 5.1)

In everyday conversation, to accelerate means to speed up. The accelerator in a car can in fact cause it to speed up. The greater the acceleration, the greater the change in velocity over a given time. The formal definition of acceleration is consistent with these notions, but more inclusive.

Average Acceleration

Average Acceleration is the rate at which velocity changes,

a-=ΔvΔt=vfv0tft0,a-=ΔvΔt=vfv0tft0, size 12{ { bar {a}}= { {Δv} over {Δt} } = { {v"" lSub { size 8{f} } - v rSub { size 8{0} } } over {t rSub { size 8{f} } - t rSub { size 8{0} } } } } {}
2.10

where a-a- size 12{ { bar {a}}} {} is average acceleration, vv size 12{v} {} is velocity, and tt size 12{t} {} is time. (The bar over the aa size 12{a} {} means average acceleration.)

Because acceleration is velocity in m/s divided by time in s, the SI units for acceleration are m/s2m/s2 size 12{"m/s" rSup { size 8{2} } } {}, meters per second squared or meters per second per second, which literally means by how many meters per second the velocity changes every second.

Recall that velocity is a vector—it has both magnitude and direction. This means that a change in velocity can be a change in magnitude (or speed), but it can also be a change in direction. For example, if a car turns a corner at constant speed, it is accelerating because its direction is changing. The quicker you turn, the greater the acceleration. So there is an acceleration when velocity changes either in magnitude (an increase or decrease in speed) or in direction, or both.

Acceleration as a Vector

Acceleration is a vector in the same direction as the change in velocity, ΔvΔv size 12{Dv} {}. Since velocity is a vector, it can change either in magnitude or in direction. Acceleration is therefore a change in either speed or direction, or both.

Keep in mind that although acceleration is in the direction of the change in velocity, it is not always in the direction of motion. When an object's acceleration is in the same direction of its motion, the object will speed up. However, when an object's acceleration is opposite to the direction of its motion, the object will slow down. Speeding up and slowing down should not be confused with a positive and negative acceleration. The next two examples should help to make this distinction clear.

A subway train arriving at a station. A velocity vector arrow points along the track away from the train. An acceleration vector arrow points along the track toward the train.
Figure 2.13 A subway train in Sao Paulo, Brazil, decelerates as it comes into a station. It is accelerating in a direction opposite to its direction of motion. (credit: Yusuke Kawasaki, Flickr)

Making Connections: Car Motion

The gray arrows are organized in 5 horizontal rows and labeled on the left edge as A, B, C, D, E. The left edge of the arrows on each row line up in a vertical line as well as the right edge of each row of arrows (the sum of all are the same length). The arrows in each of the rows are as follows: Row A's four arrows are pointing right: shortest, slightly longer, longer yet, longest. Row B's four arrows are pointing right: longest, slightly shorter, shorter yet, shortest (same direction but reverse order of sizes of A's row. Row C's arrows are pointing right: There are six arrows all the same as the shortest size of the arrows in row A and B (but the total length of all arrows and spaces is the same in all rows). Row D's four arrows are pointing left: longest, slightly shorter, shorter yet, shortest (same size but opposite direction of B's row). Row E's four arrows are pointing left: shortest, slightly longer, longer yet, longest (same size but opposite direction of A's row).
Figure 2.14 Above are arrows representing the motion of five cars (A–E). In all five cases, the positive direction should be considered to the right of the page.

Consider the acceleration and velocity of each car in terms of its direction of travel.

Figure 2.15 Car A is speeding up.

Because the positive direction is considered to the right of the paper, Car A is moving with a positive velocity. Because it is speeding up while moving with a positive velocity, its acceleration is also considered positive.

The diagram shows red car facing right in a road. Below the car is a B with four arrows pointing to the right: longest, slightly shorter, shorter yet, shortest.
Figure 2.16 Car B is slowing down.

Because the positive direction is considered to the right of the paper, Car B is also moving with a positive velocity. However, because it is slowing down while moving with a positive velocity, its acceleration is considered negative. (This can be viewed in a mathematical manner as well. If the car was originally moving with a velocity of +25 m/s, it is finishing with a speed less than that, like +5 m/s. Because the change in velocity is negative, the acceleration will be as well.)

The diagram shows red car facing right in a road. Below the car is a C with six arrows pointing to the right. The arrows are all the same as the shortest size of the arrows in row A and B.
Figure 2.17 Car C has a constant speed.

Because the positive direction is considered to the right of the paper, Car C is moving with a positive velocity. Because all arrows are of the same length, this car is not changing its speed. As a result, its change in velocity is zero, and its acceleration must be zero as well.

The diagram shows red car facing left in a road. Below the car is a D with four arrows pointing to the left: shortest, slightly longer, longer yet, longest (going from right to left – the same direction as the car).
Figure 2.18 Car D is speeding up in the opposite direction of Cars A, B, C.

Because the car is moving opposite to the positive direction, Car D is moving with a negative velocity. Because it is speeding up while moving in a negative direction, its acceleration is negative as well.

The diagram shows red car facing left in a road. Below the car is an E with four arrows pointing to the left: longest, slightly shorter, shorter yet, shortest (going from right to left – the same direction as the car).
Figure 2.19 Car E is slowing down in the same direction as Car D and opposite of Cars A, B, C.

Because it is moving opposite to the positive direction, Car E is moving with a negative velocity as well. However, because it is slowing down while moving in a negative direction, its acceleration is actually positive. As in example B, this may be more easily understood in a mathematical sense. The car is originally moving with a large negative velocity (−25 m/s) but slows to a final velocity that is less negative (−5 m/s). This change in velocity, from −25 m/s to −5 m/s, is actually a positive change ( v f v i =5 m/s 25 m/s v f v i =5 m/s 25 m/s of 20 m/s. Because the change in velocity is positive, the acceleration must also be positive.

Making Connection - Illustrative Example

The three graphs below are labeled A, B, and C. Each one represents the position of a moving object plotted against time.

The three graphs are labeled A, B, and C moving from left to right. The left, vertical axis on each graph is labeled Position. The bottom, horizontal axis is labeled Time. In graph A, the green curve line begins at the origin and starts horizontally with an increasing slope until the line is nearly vertical. In graph B, the green curve line begins at the origin and starts vertically with a decreasing slope until the line is nearly horizontal. In graph C, the green curve begins near the top of the Position axis and starts horizontally until it is nearly vertical at the end of the Time axis.
Figure 2.20 Three position and time graphs: A, B, and C.

As we did in the previous example, let's consider the acceleration and velocity of each object in terms of its direction of travel.

The green curve line begins at the origin and starts horizontally with an increasing slope until the line is nearly vertical.
Figure 2.21 Graph A of Position (y axis) vs. Time (x axis).

Object A is continually increasing its position in the positive direction. As a result, its velocity is considered positive.

Graph A above has a gray rectangle indicating about ½ of the horizontal Time and 1/5th of the vertical Position. The gray rectangle surrounds the green line on the bottom left of the graph and is longer than it is tall. A much larger green rectangle surrounds the last portion of the green curve on the right half of the graph. The width is only slightly less than the width of the gray rectangle and has about five times the height of the gray rectangle.
Figure 2.22 Breakdown of Graph A into two separate sections.

During the first portion of time (shaded grey) the position of the object does not change much, resulting in a small positive velocity. During a later portion of time (shaded green) the position of the object changes more, resulting in a larger positive velocity. Because this positive velocity is increasing over time, the acceleration of the object is considered positive.

In graph B, the green curve line begins at the origin and starts vertically with a decreasing slope until the line is nearly horizontal.
Figure 2.23 Graph B of Position (y axis) vs. Time (x axis).

As in case A, Object B is continually increasing its position in the positive direction. As a result, its velocity is considered positive.

Graph B above has a gray rectangle indicating about ½ of the horizontal Time and 4/5th of the vertical Position in the left half of the graph. The gray rectangle surrounds the green line and is taller than it is wide in the top half of the right side of the graph. A much smaller green rectangle surrounds the last portion of the green curve. The width is only slightly less than the width of the gray rectangle but has very little height (about 1/5th of the gray rectangle).
Figure 2.24 Breakdown of Graph B into two separate sections.

During the first portion of time (shaded grey) the position of the object changes a large amount, resulting in a large positive velocity. During a later portion of time (shaded green) the position of the object does not change as much, resulting in a smaller positive velocity. Because this positive velocity is decreasing over time, the acceleration of the object is considered negative.

In graph C, the green curve begins near the top of the Position axis and starts horizontally until it is nearly vertical at the end of the Time axis.
Figure 2.25 Graph C of Position (y axis) vs. Time (x axis).

Object C is continually decreasing its position in the positive direction. As a result, its velocity is considered negative.

Graph C above has a gray rectangle indicating about ½ of the horizontal Time and 1/5th of the vertical Position. The gray rectangle surrounds the green line and is much shorter than it is wide and starts at the top of the graph. A much larger green rectangle surrounds the last portion of the green curve on the right half of the graph. The width is only slightly less than the width of the gray rectangle and has about five times the height of the gray rectangle reaching to the Time axis.
Figure 2.26 Breakdown of Graph C into two separate sections.

During the first portion of time (shaded grey) the position of the object does not change a large amount, resulting in a small negative velocity. During a later portion of time (shaded green) the position of the object changes a much larger amount, resulting in a larger negative velocity. Because the velocity of the object is becoming more negative during the time period, the change in velocity is negative. As a result, the object experiences a negative acceleration.

Example 2.1 Calculating Acceleration: A Racehorse Leaves the Gate

A racehorse coming out of the gate accelerates from rest to a velocity of 15.0 m/s due west in 1.80 s. What is its average acceleration?

Two racehorses running toward the left.
Figure 2.27 (credit: Jon Sullivan, PD Photo.org)

Strategy

First we draw a sketch and assign a coordinate system to the problem. This is a simple problem, but it always helps to visualize it. Notice that we assign east as positive and west as negative. Thus, in this case, we have negative velocity.

An acceleration vector arrow pointing west, in the negative x direction, labeled with a equals question mark. A velocity vector arrow also pointing toward the left, with initial velocity labeled as 0 and final velocity labeled as negative fifteen point 0 meters per second.
Figure 2.28

We can solve this problem by identifying ΔvΔv and ΔtΔt from the given information and then calculating the average acceleration directly from the equation a-=ΔvΔt=vfv0tft0a-=ΔvΔt=vfv0tft0.

Solution

1. Identify the knowns. v0=0v0=0, vf=15.0 m/svf=15.0 m/s (the minus sign indicates direction toward the west), Δt=1.80 sΔt=1.80 s.

2. Find the change in velocity. Since the horse is going from zero to 15.0 m/s15.0 m/s size 12{ - "15" "." 0`"m/s"} {}, its change in velocity equals its final velocity: Δv=vf=15.0 m/sΔv=vf=15.0 m/s.

3. Plug in the known values (ΔvΔv and ΔtΔt) and solve for the unknown a-a-.

a-=ΔvΔt=15.0 m/s 1.80 s=8.33 m/s2.a-=ΔvΔt=15.0 m/s 1.80 s=8.33 m/s2.
2.11

Discussion

The minus sign for acceleration indicates that acceleration is toward the west. An acceleration of 8.33 m/s28.33 m/s2 due west means that the horse increases its velocity by 8.33 m/s due west each second, that is, 8.33 meters per second per second, which we write as 8.33 m/s28.33 m/s2 size 12{8 "." "33"`"m/s" rSup { size 8{2} } } {}. This is truly an average acceleration, because the ride is not smooth. We shall see later that an acceleration of this magnitude would require the rider to hang on with a force nearly equal to his weight.

Instantaneous Acceleration

Instantaneous acceleration aa, or the acceleration at a specific instant in time, is obtained by the same process as discussed for instantaneous velocity in Time, Velocity, and Speed—that is, by considering an infinitesimally small interval of time. How do we find instantaneous acceleration using only algebra? The answer is that we choose an average acceleration that is representative of the motion. Figure 2.29 shows graphs of instantaneous acceleration versus time for two very different motions. In Figure 2.29(a), the acceleration varies slightly and the average over the entire interval is nearly the same as the instantaneous acceleration at any time. In this case, we should treat this motion as if it had a constant acceleration equal to the average (in this case about 1.8 m/s21.8 m/s2). In Figure 2.29(b), the acceleration varies drastically over time. In such situations it is best to consider smaller time intervals and choose an average acceleration for each. For example, we could consider motion over the time intervals from 0 to 1.0 s and from 1.0 to 3.0 s as separate motions with accelerations of +3.0 m/s2+3.0 m/s2 and –2.0 m/s2–2.0 m/s2, respectively.

Line graphs of instantaneous acceleration in meters per second per second versus time in seconds. The line on graph (a) shows slight variation above and below an average acceleration of about 1 point 8 meters per second per second. The line on graph (b) shows great variation over time, with instantaneous acceleration constant at 3 point 0 meters per second per second for 1 second, then dropping to negative 2 point 0 meters per second per second for the next 2 seconds, and then rising again, and so forth.
Figure 2.29 Graphs of instantaneous acceleration versus time for two different one-dimensional motions. (a) Here acceleration varies only slightly and is always in the same direction, since it is positive. The average over the interval is nearly the same as the acceleration at any given time. (b) Here the acceleration varies greatly, perhaps representing a package on a post office conveyor belt that is accelerated forward and backward as it bumps along. It is necessary to consider small time intervals (such as from 0 to 1.0 s) with constant or nearly constant acceleration in such a situation.

The next several examples consider the motion of the subway train shown in Figure 2.30. In (a) the shuttle moves to the right, and in (b) it moves to the left. The examples are designed to further illustrate aspects of motion and to illustrate some of the reasoning that goes into solving problems.

In part (a), a subway train moves from left to right from an initial position of x equals 4 point 7 kilometers to a final position of x equals 6 point 7 kilometers, with a displacement of 2 point 0 kilometers. In part (b), the train moves toward the left, from an initial position of 5 point 25 kilometers to a final position of 3 point 75 kilometers.
Figure 2.30 One-dimensional motion of a subway train considered in Example 2.2, Example 2.3, Example 2.4, Example 2.5, Example 2.6, and Example 2.7. Here we have chosen the xx-axis so that + means to the right and means to the left for displacements, velocities, and accelerations. (a) The subway train moves to the right from x0x0 to xfxf. Its displacement ΔxΔx is +2.0 km. (b) The train moves to the left from x0x0 to xfxf size 12{ { {x}} sup { ' } rSub { size 8{f} } } {}. Its displacement ΔxΔx size 12{Δx'} {} is 1.5 km1.5 km. (Note that the prime symbol (′) is used simply to distinguish between displacement in the two different situations. The distances of travel and the size of the cars are on different scales to fit everything into the diagram.)

Example 2.2 Calculating Displacement: A Subway Train

What are the magnitude and sign of displacements for the motions of the subway train shown in parts (a) and (b) of Figure 2.30?

Strategy

A drawing with a coordinate system is already provided, so we don't need to make a sketch, but we should analyze it to make sure we understand what it is showing. Pay particular attention to the coordinate system. To find displacement, we use the equation Δx=xfx0Δx=xfx0 size 12{Δx=x rSub { size 8{f} } - x rSub { size 8{0} } } {}. This is straightforward since the initial and final positions are given.

Solution

1. Identify the knowns. In the figure we see that xf=6.70 kmxf=6.70 km and x0=4.70 kmx0=4.70 km for part (a), and xf=3.75 kmxf=3.75 km and x0=5.25 kmx0=5.25 km for part (b).

2. Solve for displacement in part (a).

Δ x = x f x 0 = 6 . 70 km 4 . 70 km = + 2 . 00 km Δ x = x f x 0 = 6 . 70 km 4 . 70 km = + 2 . 00 km size 12{Δx=x rSub { size 8{f} } - x rSub { size 8{0} } =6 "." "70 km" - 4 "." "70 km""=+"2 "." "00 km"} {}
2.12

3. Solve for displacement in part (b).

Δx = x f x 0 = 3.75 km 5.25 km = 1.50 km Δx = x f x 0 = 3.75 km 5.25 km = 1.50 km size 12{Δx'= { {x}} sup { ' } rSub { size 8{f} } - { {x}} sup { ' } rSub { size 8{0} } =3 "." "75 km" - 5 "." "25 km"= - 1 "." "50 km"} {}
2.13

Discussion

The direction of the motion in (a) is to the right and therefore its displacement has a positive sign, whereas motion in (b) is to the left and thus has a minus sign.

Example 2.3 Comparing Distance Traveled with Displacement: A Subway Train

What are the distances traveled for the motions shown in parts (a) and (b) of the subway train in Figure 2.30?

Strategy

To answer this question, think about the definitions of distance and distance traveled, and how they are related to displacement. Distance between two positions is defined to be the magnitude of displacement, which was found in Example 2.2. Distance traveled is the total length of the path traveled between the two positions. (See Displacement.) In the case of the subway train shown in Figure 2.30, the distance traveled is the same as the distance between the initial and final positions of the train.

Solution

1. The displacement for part (a) was +2.00 km. Therefore, the distance between the initial and final positions was 2.00 km, and the distance traveled was 2.00 km.

2. The displacement for part (b) was −1.5 km.−1.5 km. Therefore, the distance between the initial and final positions was 1.50 km, and the distance traveled was 1.50 km.

Discussion

Distance is a scalar. It has magnitude but no sign to indicate direction.

Example 2.4 Calculating Acceleration: A Subway Train Speeding Up

Suppose the train in Figure 2.30(a) accelerates from rest to 30.0 km/h in the first 20.0 s of its motion. What is its average acceleration during that time interval?

Strategy

It is worth it at this point to make a simple sketch:

A point represents the initial velocity of 0 kilometers per second. Below the point is a velocity vector arrow pointing to the right, representing the final velocity of thirty point zero kilometers per hour. Below the velocity vector is an acceleration vector arrow labeled a equals question mark.
Figure 2.31

This problem involves three steps. First we must determine the change in velocity, then we must determine the change in time, and finally we use these values to calculate the acceleration.

Solution

1. Identify the knowns. v0=0v0=0 size 12{v rSub { size 8{0} } =0} {} (the trains starts at rest), vf=30.0 km/hvf=30.0 km/h size 12{v rSub { size 8{f} } ="30" "." "0 km/h"} {}, and Δt=20.0 sΔt=20.0 s size 12{Δt="20" "." "0 s"} {}.

2. Calculate ΔvΔv size 12{Δv} {}. Since the train starts from rest, its change in velocity is Δv=+30.0 km/hΔv=+30.0 km/h size 12{Δv"=+""30" "." 0`"km/h"} {}, where the plus sign means velocity to the right.

3. Plug in known values and solve for the unknown, a-a- size 12{ { bar {a}}} {}.

a - = Δv Δt = + 30.0 km/h 20 . 0 s a - = Δv Δt = + 30.0 km/h 20 . 0 s size 12{ { bar {a}}= { {Δv} over {Δt} } = { {+"30" "." 0`"km/h"} over {"20" "." 0`s} } } {}
2.14

4. Since the units are mixed (we have both hours and seconds for time), we need to convert everything into SI units of meters and seconds. (See Physical Quantities and Units for more guidance.)

a - = + 30 km/h 20.0 s 10 3 m 1 km 1 h 3600 s = 0 . 417 m/s 2 a - = + 30 km/h 20.0 s 10 3 m 1 km 1 h 3600 s = 0 . 417 m/s 2 size 12{ { bar {a}}= left ( { {+"30 km/h"} over {"20" "." "0 s"} } right ) left ( { {"10" rSup { size 8{3} } " m"} over {"1 km"} } right ) left ( { {"1 h"} over {"3600 s"} } right )=0 "." "417 m/s" rSup { size 8{2} } } {}
2.15

Discussion

The plus sign means that acceleration is to the right. This is reasonable because the train starts from rest and ends up with a velocity to the right (also positive). So acceleration is in the same direction as the change in velocity, as is always the case.

Example 2.5 Calculate Acceleration: A Subway Train Slowing Down

Now suppose that at the end of its trip, the train in Figure 2.30(a) slows to a stop from a speed of 30.0 km/h in 8.00 s. What is its average acceleration while stopping?

Strategy

A velocity vector arrow pointing toward the right with initial velocity of thirty point zero kilometers per hour and final velocity of 0. An acceleration vector arrow pointing toward the left, labeled a equals question mark.
Figure 2.32

In this case, the train is decelerating and its acceleration is negative because it is toward the left. As in the previous example, we must find the change in velocity and the change in time and then solve for acceleration.

Solution

1. Identify the knowns. v0=30.0 km/hv0=30.0 km/h, vf=0 km/hvf=0 km/h (the train is stopped, so its velocity is 0), and Δt=8.00 sΔt=8.00 s.

2. Solve for the change in velocity, ΔvΔv size 12{Δv} {}.

Δ v = v f v 0 = 0 30 . 0 km/h = 30 .0 km/h Δ v = v f v 0 = 0 30 . 0 km/h = 30 .0 km/h size 12{Δv=v rSub { size 8{f} } - v rSub { size 8{0} } =0 - "30" "." "0 km/h"= - "30" "." "0 km/h"} {}
2.16

3. Plug in the knowns, ΔvΔv size 12{Δv} {} and ΔtΔt, and solve for a-a-.

a - = Δv Δt = 30 . 0 km/h 8 . 00 s a - = Δv Δt = 30 . 0 km/h 8 . 00 s size 12{ { bar {a}}= { {Δv} over {Δt} } = { { - "30" "." "0 km/h"} over {8 "." "00 s"} } } {}
2.17

4. Convert the units to meters and seconds.

a - = Δv Δt = 30.0 km/h 8.00 s 10 3 m 1 km 1 h 3600 s = −1.04 m/s 2 . a - = Δv Δt = 30.0 km/h 8.00 s 10 3 m 1 km 1 h 3600 s = −1.04 m/s 2 . size 12{ { bar {a}}= { {Δv} over {Δt} } = left ( { { - "30" "." "0 km/h"} over {8 "." "00 s"} } right ) left ( { {"10" rSup { size 8{3} } " m"} over {"1 km"} } right ) left ( { {"1 h"} over {"3600 s"} } right )= - 1 "." "04 m/s" rSup { size 8{2} } "." } {}
2.18

Discussion

The minus sign indicates that acceleration is to the left. This sign is reasonable because the train initially has a positive velocity in this problem, and a negative acceleration would oppose the motion. Again, acceleration is in the same direction as the change in velocity, which is negative here. This acceleration can be called a deceleration because it has a direction opposite to the velocity.

The graphs of position, velocity, and acceleration vs. time for the trains in Example 2.4 and Example 2.5 are displayed in Figure 2.33. (We have taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.)

Three graphs. The first is a line graph of position in meters versus time in seconds. The line begins at the origin and has a concave up shape from time equals zero to time equals twenty seconds. It is straight with a positive slope from twenty seconds to forty seconds. It is then convex up from forty to fifty seconds. The second graph is a line graph of velocity in meters per second versus time in seconds. The line is straight with a positive slope beginning at the origin from 0 to twenty seconds. It is flat from twenty to forty seconds. From forty to fifty seconds the line is straight with a negative slope back down to a velocity of 0. The third graph is a line graph of acceleration in meters per second per second versus time in seconds. The line is flat with a positive constant acceleration from zero to twenty seconds. The line then drops to an acceleration of 0 from twenty to forty seconds. The line drops again to a negative acceleration from forty to fifty seconds.
Figure 2.33 (a) Position of the train over time. Notice that the train's position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a constant rate. (b) Velocity of the train over time. The train's velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey.

Example 2.6 Calculating Average Velocity: The Subway Train

What is the average velocity of the train in part b of Example 2.2, and shown again below, if it takes 5.00 min to make its trip?

The train moves toward the left, from an initial position of 5 point 25 kilometers to a final position of 3 point 75 kilometers.
Figure 2.34

Strategy

Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement.

Solution

1. Identify the knowns. xf=3.75 kmxf=3.75 km, x0=5.25 kmx0=5.25 km, Δt=5.00 minΔt=5.00 min.

2. Determine displacement, ΔxΔx. We found ΔxΔx to be 1.5 km1.5 km in Example 2.2.

3. Solve for average velocity.

v - = Δ x Δ t = 1.50 km 5.00 min v - = Δ x Δ t = 1.50 km 5.00 min size 12{ { bar {v}}= { {Δ { {x}} sup { ' }} over {Δt} } = { { - 1 "." "50 km"} over {5 "." "00 min"} } } {}
2.19

4. Convert units.

v - = Δ x Δ t = 1 . 50 km 5 . 00 min 60 min 1 h = 18 .0 km/h v - = Δ x Δ t = 1 . 50 km 5 . 00 min 60 min 1 h = 18 .0 km/h size 12{ { bar {v}}= { {Δx'} over {Δt} } = left ( { { - 1 "." "50"`"km"} over {5 "." "00"`"min"} } right ) left ( { {"60"`"min"} over {1`h} } right )= - "18" "." 0`"km/h"} {}
2.20

Discussion

The negative velocity indicates motion to the left.

Example 2.7 Calculating Deceleration: The Subway Train

Finally, suppose the train in Figure 2.34 slows to a stop from a velocity of 20.0 km/h in 10.0 s. What is its average acceleration?

Strategy

Once again, let's draw a sketch:

A velocity vector arrow pointing to the left with initial velocity of negative twenty point 0 kilometers per hour and a final velocity of 0. An acceleration vector arrow pointing toward the right, labeled a equals question mark.
Figure 2.35

As before, we must find the change in velocity and the change in time to calculate average acceleration.

Solution

1. Identify the knowns. v0=20 km/hv0=20 km/h, vf=0 km/hvf=0 km/h, Δt=10.0 sΔt=10.0 s.

2. Calculate ΔvΔv size 12{Δv} {}. The change in velocity here is actually positive, since

Δv=vfv0=020 km/h=+20 km/h.Δv=vfv0=020 km/h=+20 km/h. size 12{Δv=v rSub { size 8{f} } - v rSub { size 8{0} } =0 - left ( - "20 km/h" right )"=+""20 km/h"} {}
2.21

3. Solve for a-a- size 12{ { bar {a}}} {}.

a - = Δv Δt = + 20 .0 km/h 10 . 0 s a - = Δv Δt = + 20 .0 km/h 10 . 0 s
2.22

4. Convert units.

a - = + 20 . 0 km/h 10 . 0 s 10 3 m 1 km 1 h 3600 s = + 0 .556 m /s 2 a - = + 20 . 0 km/h 10 . 0 s 10 3 m 1 km 1 h 3600 s = + 0 .556 m /s 2
2.23

Discussion

The plus sign means that acceleration is to the right. This is reasonable because the train initially has a negative velocity (to the left) in this problem and a positive acceleration opposes the motion (and so it is to the right). Again, acceleration is in the same direction as the change in velocity, which is positive here. As in Example 2.5, this acceleration can be called a deceleration since it is in the direction opposite to the velocity.

Sign and Direction

Perhaps the most important thing to note about these examples is the signs of the answers. In our chosen coordinate system, plus means the quantity is to the right and minus means it is to the left. This is easy to imagine for displacement and velocity. But it is a little less obvious for acceleration. Most people interpret negative acceleration as the slowing of an object. This was not the case in Example 2.7, where a positive acceleration slowed a negative velocity. The crucial distinction was that the acceleration was in the opposite direction from the velocity. In fact, a negative acceleration will increase a negative velocity. For example, the train moving to the left in Figure 2.34 is sped up by an acceleration to the left. In that case, both vv size 12{v} {} and aa size 12{a} {} are negative. The plus and minus signs give the directions of the accelerations. If acceleration has the same sign as the velocity, the object is speeding up. If acceleration has the opposite sign as the velocity, the object is slowing down.

Check Your Understanding

An airplane lands on a runway traveling east. Describe its acceleration.

Solution

If we take east to be positive, then the airplane has negative acceleration, as it is accelerating toward the west. It is also decelerating: its acceleration is opposite in direction to its velocity.

PhET Explorations: Moving Man Simulation

Learn about position, velocity, and acceleration graphs. Move the little man back and forth with the mouse and plot his motion. Set the position, velocity, or acceleration and let the simulation move the man for you.

Figure 2.36
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