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College Physics for AP® Courses

2.5 Motion Equations for Constant Acceleration in One Dimension

College Physics for AP® Courses2.5 Motion Equations for Constant Acceleration in One Dimension
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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
Four men racing up a river in their kayaks.
Figure 2.37 Kinematic equations can help us describe and predict the motion of moving objects such as these kayaks racing in Newbury, England. (credit: Barry Skeates, Flickr)

Learning Objectives

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

  • Calculate displacement of an object that is not accelerating, given initial position and velocity.
  • Calculate final velocity of an accelerating object, given initial velocity, acceleration, and time.
  • Calculate displacement and final position of an accelerating object, given initial position, initial velocity, time, and acceleration.

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, or 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)

We might know that the greater the acceleration of, say, a car moving away from a stop sign, the greater the displacement in a given time. But we have not developed a specific equation that relates acceleration and displacement. In this section, we develop some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration already covered.

Notation: t, x, v, a

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is Δt=tft0Δt=tft0, taking t0=0t0=0 means that Δt=tfΔt=tf, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is, x0x0 is the initial position and v0v0 is the initial velocity. We put no subscripts on the final values. That is, tt is the final time, xx is the final position, and vv is the final velocity. This gives a simpler expression for elapsed time—now, Δt=tΔt=t. It also simplifies the expression for displacement, which is now Δx=xx0Δx=xx0. Also, it simplifies the expression for change in velocity, which is now Δv=vv0Δv=vv0. To summarize, using the simplified notation, with the initial time taken to be zero,

Δt = t Δx = x x 0 Δv = v v 0 Δt = t Δx = x x 0 Δv = v v 0
2.24

where the subscript 0 denotes an initial value and the absence of a subscript denotes a final value in whatever motion is under consideration.

We now make the important assumption that acceleration is constant. This assumption allows us to avoid using calculus to find instantaneous acceleration. Since acceleration is constant, the average and instantaneous accelerations are equal. That is,

a-=a=constant,a-=a=constant, size 12{ { bar {a}}=a="constant"} {}
2.25

so we use the symbol aa size 12{a} {} for acceleration at all times. Assuming acceleration to be constant does not seriously limit the situations we can study nor degrade the accuracy of our treatment. For one thing, acceleration is constant in a great number of situations. Furthermore, in many other situations we can accurately describe motion by assuming a constant acceleration equal to the average acceleration for that motion. Finally, in motions where acceleration changes drastically, such as a car accelerating to top speed and then braking to a stop, the motion can be considered in separate parts, each of which has its own constant acceleration.

Solving for Displacement (ΔxΔx) and Final Position (xx size 12{x} {}) from Average Velocity when Acceleration (aa size 12{a} {}) is Constant

To get our first two new equations, we start with the definition of average velocity:

v - = Δx Δt . v - = Δx Δt . size 12{ { bar {v}}= { {Δx} over {Δt} } "." } {}
2.26

Substituting the simplified notation for ΔxΔx and ΔtΔt yields

v - = x x 0 t . v - = x x 0 t . size 12{ { bar {v}}= { {x - x rSub { size 8{0} } } over {t} } "." } {}
2.27

Solving for xx size 12{x} {} yields

x = x 0 + v - t , x = x 0 + v - t , size 12{x=x rSub { size 8{0} } + { bar {v}}t" " \( "constant a" \) ,} {}
2.28

where the average velocity is

v - = v 0 + v 2 ( constant a ) . v - = v 0 + v 2 ( constant a ) . size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } " " \( "constant "a \) "." } {}
2.29

The equation v-=v0+v2v-=v0+v2 size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } } {} reflects the fact that, when acceleration is constant, vv size 12{v} {} is just the simple average of the initial and final velocities. For example, if you steadily increase your velocity (that is, with constant acceleration) from 30 to 60 km/h, then your average velocity during this steady increase is 45 km/h. Using the equation v-=v0+v2v-=v0+v2 size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } } {} to check this, we see that

v - = v 0 + v 2 = 30 km/h + 60 km/h 2 = 45 km/h, v - = v 0 + v 2 = 30 km/h + 60 km/h 2 = 45 km/h, size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } = { {"30 km/h"+"60 km/h"} over {2} } ="45 km/h,"} {}
2.30

which seems logical.

Example 2.8 Calculating Displacement: How Far does the Jogger Run?

A jogger runs down a straight stretch of road with an average velocity of 4.00 m/s for 2.00 min. What is his final position, taking his initial position to be zero?

Strategy

Draw a sketch.

Velocity vector arrow labeled v equals 4 point zero zero meters per second over an x axis displaying initial and final positions. Final position is labeled x equals question mark.
Figure 2.38

The final position xx size 12{x} {} is given by the equation

x=x0+v-t.x=x0+v-t. size 12{x=x rSub { size 8{0} } + { bar {v}}t} {}
2.31

To find xx size 12{x} {}, we identify the values of x0x0 size 12{x rSub { size 8{0} } } {}, v-v- size 12{ { bar {v}}} {}, and tt size 12{t} {} from the statement of the problem and substitute them into the equation.

Solution

1. Identify the knowns. v-=4.00 m/sv-=4.00 m/s, Δt=2.00 minΔt=2.00 min size 12{Δt=2 "." "00 min"} {}, and x0=0 mx0=0 m size 12{x rSub { size 8{0} } ="0 m"} {}.

2. Enter the known values into the equation.

x = x 0 + v - t = 0 + 4 . 00 m/s 120 s = 480 m x = x 0 + v - t = 0 + 4 . 00 m/s 120 s = 480 m size 12{x=x rSub { size 8{0} } + { bar {v}}t=0+ left (4 "." "00 m/s" right ) left ("120 s" right )="480 m"} {}
2.32

Discussion

Velocity and final displacement are both positive, which means they are in the same direction.

The equation x=x0+v-tx=x0+v-t size 12{x=x rSub { size 8{0} } + { bar {v}}t} {} gives insight into the relationship between displacement, average velocity, and time. It shows, for example, that displacement is a linear function of average velocity. (By linear function, we mean that displacement depends on v-v- size 12{ { bar {v}}} {} rather than on v-v- size 12{ { bar {v}}} {} raised to some other power, such as v-2v-2 size 12{ { bar {v}} rSup { size 8{2} } } {}. When graphed, linear functions look like straight lines with a constant slope.) On a car trip, for example, we will get twice as far in a given time if we average 90 km/h than if we average 45 km/h.

Line graph showing displacement in meters versus average velocity in meters per second. The line is straight with a positive slope. Displacement x increases linearly with increase in average velocity v.
Figure 2.39 There is a linear relationship between displacement and average velocity. For a given time tt size 12{t} {}, an object moving twice as fast as another object will move twice as far as the other object.

Solving for Final Velocity

We can derive another useful equation by manipulating the definition of acceleration.

a = Δv Δt a = Δv Δt
2.33

Substituting the simplified notation for ΔvΔv and ΔtΔt gives us

a = v v 0 t ( constant a ) . a = v v 0 t ( constant a ) . size 12{a= { {v - v rSub { size 8{0} } } over {t} } " " \( "constant "a \) "." } {}
2.34

Solving for vv size 12{v} {} yields

v = v 0 + at ( constant a ) . v = v 0 + at ( constant a ) . size 12{v=v rSub { size 8{0} } + ital "at"" " \( "constant "a \) "." } {}
2.35

Example 2.9 Calculating Final Velocity: An Airplane Slowing Down after Landing

An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1.50 m/s21.50 m/s2 size 12{1 "." "50 m/s" rSup { size 8{2} } } {} for 40.0 s. What is its final velocity?

Strategy

Draw a sketch. We draw the acceleration vector in the direction opposite the velocity vector because the plane is decelerating.

Velocity vector arrow pointing toward the right in the positive x direction. Initial velocity equals seventy meters per second. Final velocity equals question mark. An acceleration vector arrow pointing toward the left labeled a equals negative 1 point 50 meters per second squared.
Figure 2.40

Solution

1. Identify the knowns. v 0 =70.0 m/s v 0 =70.0 m/s size 12{Δv="70" "." "0 m/s"} {}, a=1.50 m/s2a=1.50 m/s2 size 12{a= - 1 "." "50 m/s" rSup { size 8{2} } } {}, t=40.0st=40.0s.

2. Identify the unknown. In this case, it is final velocity, vfvf size 12{v rSub { size 8{f} } } {}.

3. Determine which equation to use. We can calculate the final velocity using the equation v=v0+atv=v0+at size 12{v=v rSub { size 8{0} } + ital "at"} {}.

4. Plug in the known values and solve.

v = v 0 + at = 70 . 0 m/s + 1 . 50 m/s 2 40 . 0 s = 10 . 0 m/s v = v 0 + at = 70 . 0 m/s + 1 . 50 m/s 2 40 . 0 s = 10 . 0 m/s size 12{v=v rSub { size 8{0} } + ital "at"="70" "." "0 m/s"+ left ( - 1 "." "50 m/s" rSup { size 8{2} } right ) left ("40" "." "0 s" right )="10" "." "0 m/s"} {}
2.36

Discussion

The final velocity is much less than the initial velocity, as desired when slowing down, but still positive. With jet engines, reverse thrust could be maintained long enough to stop the plane and start moving it backward. That would be indicated by a negative final velocity, which is not the case here.

An airplane moving toward the right at two points in time. At time equals 0 the velocity vector arrow points toward the right and is labeled seventy meters per second. The acceleration vector arrow points toward the left and is labeled negative 1 point 5 meters per second squared. At time equals forty seconds, the velocity arrow is shorter, points toward the right, and is labeled ten meters per second. The acceleration vector arrow is still pointing toward the left and is labeled a equals negative 1 point 5 meters per second squared.
Figure 2.41 The airplane lands with an initial velocity of 70.0 m/s and slows to a final velocity of 10.0 m/s before heading for the terminal. Note that the acceleration is negative because its direction is opposite to its velocity, which is positive.

In addition to being useful in problem solving, the equation v=v0+atv=v0+at size 12{v=v rSub { size 8{0} } + ital "at"} {} gives us insight into the relationships among velocity, acceleration, and time. From it we can see, for example, that

  • final velocity depends on how large the acceleration is and how long it lasts
  • if the acceleration is zero, then the final velocity equals the initial velocity (v=v0)(v=v0) size 12{ \( v=v rSub { size 8{0} } \) } {}, as expected (i.e., velocity is constant)
  • if aa size 12{a} {} is negative, then the final velocity is less than the initial velocity

(All of these observations fit our intuition, and it is always useful to examine basic equations in light of our intuition and experiences to check that they do indeed describe nature accurately.)

Making Connections: Real-World Connection


Space shuttle blasting off at night.
Figure 2.42 The Space Shuttle Endeavor blasts off from the Kennedy Space Center in February 2010. (credit: Matthew Simantov, Flickr)

An intercontinental ballistic missile (ICBM) has a larger average acceleration than the Space Shuttle and achieves a greater velocity in the first minute or two of flight (actual ICBM burn times are classified—short-burn-time missiles are more difficult for an enemy to destroy). But the Space Shuttle obtains a greater final velocity, so that it can orbit the earth rather than come directly back down as an ICBM does. The Space Shuttle does this by accelerating for a longer time.

Solving for Final Position When Velocity is Not Constant (a0a0)

We can combine the equations above to find a third equation that allows us to calculate the final position of an object experiencing constant acceleration. We start with

v=v0+at.v=v0+at. size 12{v=v rSub { size 8{0} } + ital "at"} {}
2.37

Adding v0v0 size 12{v rSub { size 8{0} } } {} to each side of this equation and dividing by 2 gives

v 0 + v 2 = v 0 + 1 2 at . v 0 + v 2 = v 0 + 1 2 at . size 12{ { {v rSub { size 8{0} } +v} over {2} } =v rSub { size 8{0} } + { {1} over {2} } ital "at" "." } {}
2.38

Since v0+v2=v-v0+v2=v- size 12{ { {v rSub { size 8{0} } +v} over {2} } = { bar {v}}} {} for constant acceleration, then

v - = v 0 + 1 2 at . v - = v 0 + 1 2 at . size 12{ { bar {v}}=v rSub { size 8{0} } + { {1} over {2} } ital "at" "." } {}
2.39

Now we substitute this expression for v-v- size 12{ { bar {v}}} {} into the equation for displacement, x=x0+v-tx=x0+v-t size 12{x=x rSub { size 8{0} } + { bar {v}}t} {}, yielding

x = x 0 + v 0 t + 1 2 at 2 ( constant a ) . x = x 0 + v 0 t + 1 2 at 2 ( constant a ) . size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } " " \( "constant "a \) "." } {}
2.40

Example 2.10 Calculating Displacement of an Accelerating Object: Dragsters

Dragsters can achieve average accelerations of 26.0 m/s226.0 m/s2 size 12{"26" "." "0 m/s" rSup { size 8{2} } } {}. Suppose such a dragster accelerates from rest at this rate for 5.56 s. How far does it travel in this time?

Dragster accelerating down a race track.
Figure 2.43 U.S. Army Top Fuel pilot Tony “The Sarge” Schumacher begins a race with a controlled burnout. (credit: Lt. Col. William Thurmond. Photo Courtesy of U.S. Army.)

Strategy

Draw a sketch.

Acceleration vector arrow pointing toward the right in the positive x direction, labeled a equals twenty-six point 0 meters per second squared. x position graph with initial position at the left end of the graph. The right end of the graph is labeled x equals question mark.
Figure 2.44

We are asked to find displacement, which is xx if we take x0x0 size 12{x rSub { size 8{0} } } {} to be zero. (Think about it like the starting line of a race. It can be anywhere, but we call it 0 and measure all other positions relative to it.) We can use the equation x=x0+v0t+12at2x=x0+v0t+12at2 size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}once we identify v0v0 size 12{v rSub { size 8{0} } } {}, aa size 12{a} {}, and tt size 12{t} {} from the statement of the problem.

Solution

1. Identify the knowns. Starting from rest means that v0=0v0=0 size 12{v rSub { size 8{0} } =0} {}, aa size 12{a} {} is given as 26.0m/s226.0m/s2 size 12{"26" "." 0`"m/s" rSup { size 8{2} } } {} and tt size 12{t} {} is given as 5.56 s.

2. Plug the known values into the equation to solve for the unknown xx size 12{x} {}:

x = x 0 + v 0 t + 1 2 at 2 . x = x 0 + v 0 t + 1 2 at 2 . size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } "." } {}
2.41

Since the initial position and velocity are both zero, this simplifies to

x = 1 2 at 2 . x = 1 2 at 2 . size 12{x= { {1} over {2} } ital "at" rSup { size 8{2} } "." } {}
2.42

Substituting the identified values of aa size 12{a} {} and tt size 12{t} {} gives

x = 1 2 26 . 0 m/s 2 5 . 56 s 2 , x = 1 2 26 . 0 m/s 2 5 . 56 s 2 , size 12{x= { {1} over {2} } left ("26" "." "0 m/s" rSup { size 8{2} } right ) left (5 "." "56 s" right ) rSup { size 8{2} } ,} {}
2.43

yielding

x=402 m.x=402 m. size 12{x="402 m"} {}
2.44

Discussion

If we convert 402 m to miles, we find that the distance covered is very close to one quarter of a mile, the standard distance for drag racing. So the answer is reasonable. This is an impressive displacement in only 5.56 s, but top-notch dragsters can do a quarter mile in even less time than this.

What else can we learn by examining the equation x=x0+v0t+12at2?x=x0+v0t+12at2? size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {} We see that:

  • displacement depends on the square of the elapsed time when acceleration is not zero. In Example 2.10, the dragster covers only one fourth of the total distance in the first half of the elapsed time
  • if acceleration is zero, then the initial velocity equals average velocity (v0=v-v0=v- size 12{v rSub { size 8{0} } = { bar {v}}} {}) and x=x0+v0t+12at2x=x0+v0t+12at2 size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {} becomes x=x0+v0tx=x0+v0t size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t} {}

Solving for Final Velocity when Velocity Is Not Constant (a0a0)

A fourth useful equation can be obtained from another algebraic manipulation of previous equations.

If we solve v=v0+atv=v0+at size 12{v=v rSub { size 8{0} } + ital "at"} {} for tt size 12{t} {}, we get

t = v v 0 a . t = v v 0 a . size 12{t= { {v - v rSub { size 8{0} } } over {a} } "." } {}
2.45

Substituting this and v-=v0+v2v-=v0+v2 size 12{ { bar {v}}= { {v rSub { size 8{0} } +v} over {2} } } {} into x=x0+v-tx=x0+v-t size 12{x=x rSub { size 8{0} } + { bar {v}}t} {}, we get

v 2 = v 0 2 + 2a x x 0 ( constant a ) . v 2 = v 0 2 + 2a x x 0 ( constant a ) . size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a left (x - x rSub { size 8{0} } right )" " \( "constant "a \) "." } {}
2.46

Example 2.11 Calculating Final Velocity: Dragsters

Calculate the final velocity of the dragster in Example 2.10 without using information about time.

Strategy

Draw a sketch.

Acceleration vector arrow pointing toward the right, labeled twenty-six point zero meters per second squared. Initial velocity equals 0. Final velocity equals question mark.
Figure 2.45

The equation v2=v02+2a(xx0)v2=v02+2a(xx0) is ideally suited to this task because it relates velocities, acceleration, and displacement, and no time information is required.

Solution

1. Identify the known values. We know that v0=0v0=0 size 12{v rSub { size 8{0} } =0} {}, since the dragster starts from rest. Then we note that xx0=402 mxx0=402 m size 12{x - x rSub { size 8{0} } ="402 m"} {} (this was the answer in Example 2.10). Finally, the average acceleration was given to be a=26.0 m/s2a=26.0 m/s2 size 12{a="26" "." "0 m/s" rSup { size 8{2} } } {}.

2. Plug the knowns into the equation v2=v02+2a(xx0)v2=v02+2a(xx0) and solve for v.v.

v2=0+226.0 m/s2402 m.v2=0+226.0 m/s2402 m. size 12{v rSup { size 8{2} } =0+2 left ("26" "." "0 m/s" rSup { size 8{2} } right ) left ("402 m" right )} {}
2.47

Thus

v2=2.09×104m2/s2.v2=2.09×104m2/s2. size 12{v rSup { size 8{2} } =2 "." "09" times "10" rSup { size 8{4} } `m rSup { size 8{2} } "/s" rSup { size 8{2} } } {}
2.48

To get vv size 12{v} {}, we take the square root:

v=2.09×104 m2/s2=145 m/s.v=2.09×104 m2/s2=145 m/s.
2.49

Discussion

145 m/s is about 522 km/h or about 324 mi/h, but even this breakneck speed is short of the record for the quarter mile. Also, note that a square root has two values; we took the positive value to indicate a velocity in the same direction as the acceleration.

An examination of the equation v2=v02+2a(xx0)v2=v02+2a(xx0) size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a \( x - x rSub { size 8{0} } \) } {} can produce further insights into the general relationships among physical quantities:

  • The final velocity depends on how large the acceleration is and the distance over which it acts
  • For a fixed deceleration, a car that is going twice as fast doesn't simply stop in twice the distance—it takes much further to stop. (This is why we have reduced speed zones near schools.)

Putting Equations Together

In the following examples, we further explore one-dimensional motion, but in situations requiring slightly more algebraic manipulation. The examples also give insight into problem-solving techniques. The box below provides easy reference to the equations needed.

Summary of Kinematic Equations (constant aa size 12{a} {})

x = x 0 + v - t x = x 0 + v - t size 12{x=`x rSub { size 8{0} } `+` { bar {v}}t} {}
2.50
v - = v 0 + v 2 v - = v 0 + v 2 size 12{ { bar {v}}=` { {v rSub { size 8{0} } +v} over {2} } } {}
2.51
v = v 0 + at v = v 0 + at size 12{v=v rSub { size 8{0} } + ital "at"} {}
2.52
x = x 0 + v 0 t + 1 2 at 2 x = x 0 + v 0 t + 1 2 at 2 size 12{x=x rSub { size 8{0} } +v rSub { size 8{0} } t+ { {1} over {2} } ital "at" rSup { size 8{2} } } {}
2.53
v 2 = v 0 2 + 2a x x 0 v 2 = v 0 2 + 2a x x 0 size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a left (x - x rSub { size 8{0} } right )} {}
2.54

Example 2.12 Calculating Displacement: How Far Does a Car Go When Coming to a Halt?

On dry concrete, a car can decelerate at a rate of 7.00 m/s27.00 m/s2 size 12{7 "." "00 m/s" rSup { size 8{2} } } {}, whereas on wet concrete it can decelerate at only 5.00 m/s25.00 m/s2 size 12{5 "." "00 m/s" rSup { size 8{2} } } {}. Find the distances necessary to stop a car moving at 30.0 m/s (about 110 km/h) (a) on dry concrete and (b) on wet concrete. (c) Repeat both calculations, finding the displacement from the point where the driver sees a traffic light turn red, taking into account his reaction time of 0.500 s to get his foot on the brake.

Strategy

Draw a sketch.

Initial velocity equals thirty meters per second. Final velocity equals 0. Acceleration dry equals negative 7 point zero zero meters per second squared. Acceleration wet equals negative 5 point zero zero meters per second squared.
Figure 2.46

In order to determine which equations are best to use, we need to list all of the known values and identify exactly what we need to solve for. We shall do this explicitly in the next several examples, using tables to set them off.

Solution for (a)

1. Identify the knowns and what we want to solve for. We know that v0=30.0 m/sv0=30.0 m/s size 12{v rSub { size 8{0} } ="30" "." "0 m/s"} {}; v=0 v=0 size 12{v="0 "} {}; a=7.00m/s2a=7.00m/s2 (aa is negative because it is in a direction opposite to velocity). We take x0x0 to be 0. We are looking for displacement ΔxΔx, or xx0xx0 size 12{x - x rSub { size 8{0} } } {}.

2. Identify the equation that will help up solve the problem. The best equation to use is

v2=v02+2axx0.v2=v02+2axx0. size 12{v rSup { size 8{2} } =v rSub { size 8{0} } rSup { size 8{2} } +2a left (x - x rSub { size 8{0} } right )} {}
2.55

This equation is best because it includes only one unknown, xx size 12{x} {}. We know the values of all the other variables in this equation. (There are other equations that would allow us to solve for xx size 12{x} {}, but they require us to know the stopping time, tt size 12{t} {}, which we do not know. We could use them but it would entail additional calculations.)

3. Rearrange the equation to solve for xx size 12{x} {}.

x x 0 = v 2 v 0 2 2a x x 0 = v 2 v 0 2 2a
2.56

4. Enter known values.

x 0 = 0 2 30 . 0 m/s 2 2 7 . 00 m/s 2 x 0 = 0 2 30 . 0 m/s 2 2 7 . 00 m/s 2 size 12{x - 0= { {0 rSup { size 8{2} } - left ("30" "." "0 m/s" right ) rSup { size 8{2} } } over {2 left ( - 7 "." "00 m/s" rSup { size 8{2} } right )} } } {}
2.57

Thus,

x = 64 . 3 m on dry concrete . x = 64 . 3 m on dry concrete . size 12{x="64" "." "3 m on dry concrete" "." } {}
2.58

Solution for (b)

This part can be solved in exactly the same manner as Part A. The only difference is that the deceleration is 5.00 m/s25.00 m/s2 size 12{ +- 5 "." "00 m/s" rSup { size 8{2} } } {}. The result is

x wet = 90 . 0 m on wet concrete . x wet = 90 . 0 m on wet concrete . size 12{x rSub { size 8{"wet"} } ="90" "." "0 m on wet concrete" "." } {}
2.59

Solution for (c)

Once the driver reacts, the stopping distance is the same as it is in Parts A and B for dry and wet concrete. So to answer this question, we need to calculate how far the car travels during the reaction time, and then add that to the stopping time. It is reasonable to assume that the velocity remains constant during the driver's reaction time.

1. Identify the knowns and what we want to solve for. We know that v-=30.0 m/sv-=30.0 m/s; treaction=0.500streaction=0.500s; areaction=0areaction=0. We take x0reactionx0reaction to be 0. We are looking for xreactionxreaction.

2. Identify the best equation to use.

x=x0+v-tx=x0+v-t size 12{x=x rSub { size 8{0} } + { bar {v}}t} {} works well because the only unknown value is xx size 12{x} {}, which is what we want to solve for.

3. Plug in the knowns to solve the equation.

x=0+30.0 m/s0.500 s=15.0 m.x=0+30.0 m/s0.500 s=15.0 m. size 12{x=0+ left ("30" "." "0 m/s" right ) left (0 "." "500 s" right )="15" "." "0 m"} {}
2.60

This means the car travels 15.0 m while the driver reacts, making the total displacements in the two cases of dry and wet concrete 15.0 m greater than if he reacted instantly.

4. Add the displacement during the reaction time to the displacement when braking.

x braking + x reaction = x total x braking + x reaction = x total size 12{x rSub { size 8{"braking"} } +x rSub { size 8{"reaction"} } =x rSub { size 8{"total"} } } {}
2.61
  1. 64.3 m + 15.0 m = 79.3 m when dry
  2. 90.0 m + 15.0 m = 105 m when wet
Diagram showing the various braking distances necessary for stopping a car. With no reaction time considered, braking distance is 64 point 3 meters on a dry surface and 90 meters on a wet surface. With reaction time of 0 point 500 seconds, braking distance is 79 point 3 meters on a dry surface and 105 meters on a wet surface.
Figure 2.47 The distance necessary to stop a car varies greatly, depending on road conditions and driver reaction time. Shown here are the braking distances for dry and wet pavement, as calculated in this example, for a car initially traveling at 30.0 m/s. Also shown are the total distances traveled from the point where the driver first sees a light turn red, assuming a 0.500 s reaction time.

Discussion

The displacements found in this example seem reasonable for stopping a fast-moving car. It should take longer to stop a car on wet rather than dry pavement. It is interesting that reaction time adds significantly to the displacements. But more important is the general approach to solving problems. We identify the knowns and the quantities to be determined and then find an appropriate equation. There is often more than one way to solve a problem. The various parts of this example can in fact be solved by other methods, but the solutions presented above are the shortest.

Example 2.13 Calculating Time: A Car Merges into Traffic

Suppose a car merges into freeway traffic on a 200-m-long ramp. If its initial velocity is 10.0 m/s and it accelerates at 2.00 m/s22.00 m/s2 size 12{2 "." "00 m/s" rSup { size 8{2} } } {}, how long does it take to travel the 200 m up the ramp? (Such information might be useful to a traffic engineer.)

Strategy

Draw a sketch.

A line segment with ends labeled x subs zero equals zero and x = two hundred. Above the line segment, the equation t equals question mark indicates that time is unknown. Three vectors, all pointing in the direction of x equals 200, represent the other knowns and unknowns. They are labeled v sub zero equals ten point zero meters per second, v equals question mark, and a equals two point zero zero meters per second squared.
Figure 2.48

We are asked to solve for the time tt size 12{t} {}. As before, we identify the known quantities in order to choose a convenient physical relationship (that is, an equation with one unknown, tt size 12{t} {}).

Solution

1. Identify the knowns and what we want to solve for. We know that v0=10 m/sv0=10 m/s size 12{v rSub { size 8{0} } ="10 m/s"} {}; a=2.00 m/s2a=2.00 m/s2 size 12{a=2 "." "00 m/s" rSup { size 8{2} } } {}; and x=200 mx=200 m size 12{x="200 m"} {}.

2. We need to solve for tt size 12{t} {}. Choose the best equation. x=x0+v0t+12at2x=x0+v0t+12at2 works best because the only unknown in the equation is the variable tt size 12{t} {} for which we need to solve.

3. We will need to rearrange the equation to solve for tt size 12{t} {}. In this case, it will be easier to plug in the knowns first.

200 m = 0 m + 10 . 0 m/s t + 1 2 2 . 00 m/s 2 t 2 200 m = 0 m + 10 . 0 m/s t + 1 2 2 . 00 m/s 2 t 2 size 12{"200 m"="0 m"+ left ("10" "." "0 m/s" right )t+ { {1} over {2} } left (2 "." "00 m/s" rSup { size 8{2} } right )t rSup { size 8{2} } } {}
2.62

4. Simplify the equation. The units of meters (m) cancel because they are in each term. We can get the units of seconds (s) to cancel by taking t=tst=ts size 12{t=t" s"} {}, where tt size 12{t} {} is the magnitude of time and s is the unit. Doing so leaves

200=10t+t2.200=10t+t2. size 12{"200"="10"t+t rSup { size 8{2} } } {}
2.63

5. Use the quadratic formula to solve for tt size 12{t} {}.

(a) Rearrange the equation to get 0 on one side of the equation.

t 2 + 10 t 200 = 0 t 2 + 10 t 200 = 0 size 12{t rSup { size 8{2} } +"10"t - "200"=0} {}
2.64

This is a quadratic equation of the form

at2+bt+c=0,at2+bt+c=0,
2.65

where the constants are a=1.00,b=10.0,andc=200a=1.00,b=10.0,andc=200 size 12{a=1 "." "00,"`b="10" "." "0,"`"and"`c= - "200"} {}.

(b) Its solutions are given by the quadratic formula:

t = b ± b 2 4 ac 2a . t = b ± b 2 4 ac 2a .
2.66

This yields two solutions for tt size 12{t} {}, which are

t=10.0and20.0.t=10.0and20.0. size 12{t="10" "." 0``"and"`` - "20" "." 0} {}
2.67

In this case, then, the time is t=tt=t size 12{t=t} {} in seconds, or

t=10.0sand20.0s.t=10.0sand20.0s. size 12{t="10" "." 0``s`"and" - "20" "." 0`s} {}
2.68

A negative value for time is unreasonable, since it would mean that the event happened 20 s before the motion began. We can discard that solution. Thus,

t=10.0s.t=10.0s. size 12{t="10" "." 0`s} {}
2.69

Discussion

Whenever an equation contains an unknown squared, there will be two solutions. In some problems both solutions are meaningful, but in others, such as the above, only one solution is reasonable. The 10.0 s answer seems reasonable for a typical freeway on-ramp.

With the basics of kinematics established, we can go on to many other interesting examples and applications. In the process of developing kinematics, we have also glimpsed a general approach to problem solving that produces both correct answers and insights into physical relationships. Problem-Solving Basics discusses problem-solving basics and outlines an approach that will help you succeed in this invaluable task.

Making Connections: Take-Home Experiment—Breaking News

We have been using SI units of meters per second squared to describe some examples of acceleration or deceleration of cars, runners, and trains. To achieve a better feel for these numbers, one can measure the braking deceleration of a car doing a slow (and safe) stop. Recall that, for average acceleration, a-=Δv/Δta-=Δv/Δt size 12{ { bar {a}}=Δv/Δt} {}. While traveling in a car, slowly apply the brakes as you come up to a stop sign. Have a passenger note the initial speed in miles per hour and the time taken (in seconds) to stop. From this, calculate the deceleration in miles per hour per second. Convert this to meters per second squared and compare with other decelerations mentioned in this chapter. Calculate the distance traveled in braking.

Check Your Understanding

A manned rocket accelerates at a rate of 20 m/s220 m/s2 size 12{"20 m/s" rSup { size 8{2} } } {} during launch. How long does it take the rocket reach a velocity of 400 m/s?

Solution

To answer this, choose an equation that allows you to solve for time tt size 12{t} {}, given only aa size 12{a} {}, v0v0 size 12{v rSub { size 8{0} } } {}, and vv size 12{v} {}.

v = v 0 + at v = v 0 + at size 12{v=v"" lSub { size 8{0} } + ital "at"} {}
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Rearrange to solve for tt size 12{t} {}.

t = v v 0 a = 400 m/s 0 m/s 20 m/s 2 = 20 s t = v v 0 a = 400 m/s 0 m/s 20 m/s 2 = 20 s size 12{t= { {v - v"" lSub { size 8{0} } } over {a} } = { {"400 m/s" - "0 m/s"} over {"20 m/s" rSup { size 8{2} } } } ="20 s"} {}
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