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  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. 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
  12. Index

Check Your Understanding

3.1

(a) The rider’s displacement is Δx=xfx0=−1kmΔx=xfx0=−1km. (The displacement is negative because we take east to be positive and west to be negative.) (b) The distance traveled is 3 km + 2 km = 5 km. (c) The magnitude of the displacement is 1 km.

3.2

(a) Taking the derivative of x(t) gives v(t) = −6t m/s. (b) No, because time can never be negative. (c) The velocity is v(1.0 s) = −6 m/s and the speed is |v(1.0s)|=6m/s|v(1.0s)|=6m/s.

3.3

Inserting the knowns, we have
a=ΔvΔt=2.0×107m/s010−4s0=2.0×1011m/s2.a=ΔvΔt=2.0×107m/s010−4s0=2.0×1011m/s2.

3.4

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

3.5

To answer this, choose an equation that allows us to solve for time t, given only a , v0 , and v:
v=v0+at.v=v0+at.
Rearrange to solve for t:
t=vv0a=400m/s0m/s20m/s2=20s.t=vv0a=400m/s0m/s20m/s2=20s.

3.6

a=23m/s2a=23m/s2.

3.7

It takes 2.47 s to hit the water. The quantity distance traveled increases faster.

3.8
  1. The velocity function is the integral of the acceleration function plus a constant of integration. By Equation 3.18,
    v(t)=a(t)dt+C1=(510t)dt+C1=5t5t2+C1.v(t)=a(t)dt+C1=(510t)dt+C1=5t5t2+C1.
    Since v(0) = 0, we have C1 = 0; so,
    v(t)=5t5t2.v(t)=5t5t2.
  2. By Equation 3.19,
    x(t)=v(t)dt+C2=(5t5t2)dt+C2=52t253t3+C2x(t)=v(t)dt+C2=(5t5t2)dt+C2=52t253t3+C2.
    Since x(0) = 0, we have C2 = 0, and
    x(t)=52t253t3.x(t)=52t253t3.
  3. The velocity can be written as v(t) = 5t(1 – t), which equals zero at t = 0, and t = 1 s.

Conceptual Questions

1.

You drive your car into town and return to drive past your house to a friend’s house.

3.

If the bacteria are moving back and forth, then the displacements are canceling each other and the final displacement is small.

5.

Distance traveled

7.

Average speed is the total distance traveled divided by the elapsed time. If you go for a walk, leaving and returning to your home, your average speed is a positive number. Since Average velocity = Displacement/Elapsed time, your average velocity is zero.

9.

Average speed. They are the same if the car doesn’t reverse direction.

11.

No, in one dimension constant speed requires zero acceleration.

13.

A ball is thrown into the air and its velocity is zero at the apex of the throw, but acceleration is not zero.

15.

Plus, minus

17.

If the acceleration, time, and displacement are the knowns, and the initial and final velocities are the unknowns, then two kinematic equations must be solved simultaneously. Also if the final velocity, time, and displacement are the knowns then two kinematic equations must be solved for the initial velocity and acceleration.

19.

a. at the top of its trajectory; b. yes, at the top of its trajectory; c. yes

21.

Earth v=v0gt=gtv=v0gt=gt; Moon v=g6tv=vgt=g6tt=6tv=g6tv=vgt=g6tt=6t; Earth y=12gt2y=12gt2 Moon y=12g6(6t)2=12g6t2=−6(12gt2)=−6yy=12g6(6t)2=12g6t2=−6(12gt2)=−6y

Problems

25.

a. x1=(−2.0m)i^x1=(−2.0m)i^, x2=(5.0m)i^x2=(5.0m)i^; b. 7.0 m east

27.

a. t=2.0t=2.0 s; b. x(6.0)x(3.0)=−8.0(−2.0)=−6.0mx(6.0)x(3.0)=−8.0(−2.0)=−6.0m

29.

a. 150.0 s, v=156.7m/sv=156.7m/s; b. 163% the speed of sound at sea level or about Mach 2.

31.
Graph shows velocity in meters per second plotted as a function of time at seconds. Velocity starts as 10 meters per second, decreases to -30 at 0.4 seconds; increases to -10 meters at 0.6 seconds, increases to 5 at 1 second, increases to 15 at 1.6 seconds.
33.
Graph shows position plotted versus time. It starts at the origin, increases reaching maximum, and then decreases close to zero.
35.

a. v(t)=(104t)m/sv(t)=(104t)m/s; v(2 s) = 2 m/s, v(3 s) = −2 m/s; b. |v(2s)|=2m/s,|v(3s)|=2m/s|v(2s)|=2m/s,|v(3s)|=2m/s; (c) v=0m/sv=0m/s

37.

a=4.29m/s2a=4.29m/s2

39.
Graph shows acceleration in meters per second squared plotted versus time in seconds. Acceleration is 0.3 meters per second squared between 0 and 20 seconds, -0.1 meters per second squared between 20 and 50 seconds, zero between 50 and 70 seconds, -0.6 between 90 and 100 seconds.
41.

a=11.1ga=11.1g

43.

150 m

45.

a. 525 m;
b. v=180m/sv=180m/s

47.

a.

Graph is a plot of acceleration a as a function of time t. Graph is non-linear with acceleration being positive at the beginning, negative at the end, and crossing x axis between points d and e and at points e and h.


b. The acceleration has the greatest positive value at tata
c. The acceleration is zero at teandthteandth
d. The acceleration is negative at ti,tj,tk,tlti,tj,tk,tl

49.

a. a=−1.3m/s2a=−1.3m/s2;
b. v0=18m/sv0=18m/s;
c. t=13.8st=13.8s

51.

v=502.20m/sv=502.20m/s

53.

a.

Figure shows object with speed equal to 0 meters per second and acceleration equal to 2.4 meters per second squared at zero point. When time is equal to 12 seconds, acceleration remains equal to 2.4 meters per second. Speed and position of the object are unknown.


b. Knowns: a=2.40m/s2,t=12.0s,v0=0m/sa=2.40m/s2,t=12.0s,v0=0m/s, and x0=0mx0=0m;
c. x=x0+v0t+12at2=12at2=2.40m/s2(12.0s)2=172.80mx=x0+v0t+12at2=12at2=2.40m/s2(12.0s)2=172.80m, the answer seems reasonable at about 172.8 m; d. v=28.8m/sv=28.8m/s

55.

a.

Figure shows object with zero speed and unknown acceleration at the beginning. After unknown time, object reaches speed of 30 centimeters per second and is at distance of 1.8 centimeters from the starting point. Acceleration of the object at this point is unknown.


b. Knowns: v=30.0cm/s,x=1.80cmv=30.0cm/s,x=1.80cm;
c. a=250cm/s2,t=0.12sa=250cm/s2,t=0.12s;
d. yes

57.

a. 6.87 m/s2; b. x=52.26mx=52.26m

59.

a. a=8450m/s2a=8450m/s2;
b. t=0.0077st=0.0077s

61.

a. a=9.18g;a=9.18g;
b. t=6.67×10−3st=6.67×10−3s;
c. a=40.0m/s2a=4.08ga=40.0m/s2a=4.08g

63.

Knowns: x=3m,v=0m/s,v0=54m/sx=3m,v=0m/s,v0=54m/s. We want a, so we can use this equation: a=486m/s2a=486m/s2.

65.

a. a=32.58m/s2a=32.58m/s2;
b. v=161.85m/sv=161.85m/s;
c. v>vmaxv>vmax, because the assumption of constant acceleration is not valid for a dragster. A dragster changes gears and would have a greater acceleration in first gear than second gear than third gear, and so on. The acceleration would be greatest at the beginning, so it would not be accelerating at 32.6m/s232.6m/s2 during the last few meters, but substantially less, and the final velocity would be less than 162m/s162m/s.

67.

a. y=−8.23mv1=18.9m/sy=−8.23mv1=18.9m/s;
b. y=−18.9mv2=−23.8m/sy=−18.9mv2=−23.8m/s;
c. y=−32.0mv3=28.7m/sy=−32.0mv3=28.7m/s;
d. y=−47.6mv4=33.6m/sy=−47.6mv4=33.6m/s;
e. y=−65.6mv5=38.5m/sy=−65.6mv5=38.5m/s

69.

a. Knowns: a=9.8m/s2v0=−1.4m/st=1.8sy0=0ma=9.8m/s2v0=−1.4m/st=1.8sy0=0m;
b. y=y0+v0t12gt2y=v0t12gt=−1.4m/s(1.8sec)12(9.8)(1.8s)2=−18.4my=y0+v0t12gt2y=v0t12gt=−1.4m/s(1.8sec)12(9.8)(1.8s)2=−18.4m and the origin is at the rescuers, who are 18.4 m above the water.

71.

a. v2=v022g(yy0)y0=0v=0y=v022g=(4.0m/s)22(9.80)=0.82mv2=v022g(yy0)y0=0v=0y=v022g=(4.0m/s)22(9.80)=0.82m; b. to the apex v=0.41sv=0.41s times 2 to the board = 0.82 s from the board to the water y=y0+v0t12gt2y=−1.80my0=0v0=4.0m/sy=y0+v0t12gt2y=−1.80my0=0v0=4.0m/s−1.8=4.0t4.9t24.9t24.0t1.80=0−1.8=4.0t4.9t24.9t24.0t1.80=0, solution to quadratic equation gives 1.13 s; c. v2=v022g(yy0)y0=0v0=4.0m/sy=−1.80mv=7.16m/sv2=v022g(yy0)y0=0v0=4.0m/sy=−1.80mv=7.16m/s

73.

Time to the apex: t=1.12st=1.12s times 2 equals 2.24 s to a height of 2.20 m. To 1.80 m in height is an additional 0.40 m. y=y0+v0t12gt2y=−0.40my0=0v0=−11.0m/sy=y0+v0t12gt2y=−0.40my0=0v0=−11.0m/s0.40=−11.0t4.9t2or4.9t2+11.0t0.40=0y=y0+v0t12gt2y=−0.40my0=0v0=−11.0m/sy=y0+v0t12gt2y=−0.40my0=0v0=−11.0m/s0.40=−11.0t4.9t2or4.9t2+11.0t0.40=0.
Take the positive root, so the time to go the additional 0.4 m is 0.04 s. Total time is 2.24s+0.04s=2.28s2.24s+0.04s=2.28s.

75.

a. v2=v022g(yy0)y0=0v=0y=2.50mv02=2gyv0=2(9.80)(2.50)=7.0m/sv2=v022g(yy0)y0=0v=0y=2.50mv02=2gyv0=2(9.80)(2.50)=7.0m/s; b. t=0.72st=0.72s times 2 gives 1.44 s in the air

77.

a. v=70.0m/sv=70.0m/s; b. time heard after rock begins to fall: 0.75 s, time to reach the ground: 6.09 s

79.

a. A=m/s2B=m/s5/2A=m/s2B=m/s5/2;
b. v(t)=a(t)dt+C1=(ABt1/2)dt+C1=At23Bt3/2+C1v(0)=0=C1sov(t0)=At023Bt03/2v(t)=a(t)dt+C1=(ABt1/2)dt+C1=At23Bt3/2+C1v(0)=0=C1sov(t0)=At023Bt03/2;
c. x(t)=v(t)dt+C2=(At23Bt3/2)dt+C2=12At2415Bt5/2+C2x(0)=0=C2sox(t0)=12At02415Bt05/2x(t)=v(t)dt+C2=(At23Bt3/2)dt+C2=12At2415Bt5/2+C2x(0)=0=C2sox(t0)=12At02415Bt05/2

81.

a. a(t)=3.2m/s2t5.0sa(t)=1.5m/s25.0st11.0sa(t)=0m/s2t>11.0sa(t)=3.2m/s2t5.0sa(t)=1.5m/s25.0st11.0sa(t)=0m/s2t>11.0s;
b. x(t)=v(t)dt+C2=3.2tdt+C2=1.6t2+C2t5.0sx(0)=0C2=0therefore,x(2.0s)=6.4mx(t)=v(t)dt+C2=[16.01.5(t5.0)]dt+C2=16t1.5(t225.0t)+C25.0t11.0sx(5s)=1.6(5.0)2=40m=16(5.0s)1.5(5225.0(5.0))+C240=98.75+C2C2=−58.75x(7.0s)=16(7.0)1.5(7225.0(7))58.75=69mx(t)=7.0dt+C2=7t+C2t11.0sx(11.0s)=16(11)1.5(11225.0(11))58.75=109=7(11.0s)+C2C2=32mx(t)=7t+32mx11.0sx(12.0s)=7(12)+32=116mx(t)=v(t)dt+C2=3.2tdt+C2=1.6t2+C2t5.0sx(0)=0C2=0therefore,x(2.0s)=6.4mx(t)=v(t)dt+C2=[16.01.5(t5.0)]dt+C2=16t1.5(t225.0t)+C25.0t11.0sx(5s)=1.6(5.0)2=40m=16(5.0s)1.5(5225.0(5.0))+C240=98.75+C2C2=−58.75x(7.0s)=16(7.0)1.5(7225.0(7))58.75=69mx(t)=7.0dt+C2=7t+C2t11.0sx(11.0s)=16(11)1.5(11225.0(11))58.75=109=7(11.0s)+C2C2=32mx(t)=7t+32mx11.0sx(12.0s)=7(12)+32=116m

Additional Problems

83.

Take west to be the positive direction.
1st plane: ν=600km/hν=600km/h
2nd plane ν=667.0km/hν=667.0km/h

85.

a=vv0tt0a=vv0tt0, t=0,a=3.4cm/sv04s=1.2cm/s2v0=8.2cm/st=0,a=3.4cm/sv04s=1.2cm/s2v0=8.2cm/s v=v0+at=8.2+1.2tv=v0+at=8.2+1.2t; v=−7.0cm/sv=−1.0cm/sv=−7.0cm/sv=−1.0cm/s

87.

a=−3m/s2a=−3m/s2

89.

a.
v=8.7×105m/sv=8.7×105m/s;
b. t=7.8×108st=7.8×108s

91.

1km=v0(80.0s)+12a(80.0)21km=v0(80.0s)+12a(80.0)2; 2km=v0(200.0)+12a(200.0)22km=v0(200.0)+12a(200.0)2 solve simultaneously to get a=0.12400.0km/s2a=0.12400.0km/s2 and v0=0.014167km/sv0=0.014167km/s, which is 51.0km/h51.0km/h. Velocity at the end of the trip is v=21.0km/hv=21.0km/h.

93.

a=−0.9m/s2a=−0.9m/s2

95.

Equation for the speeding car: This car has a constant velocity, which is the average velocity, and is not accelerating, so use the equation for displacement with x0=0x0=0:x=x0+vt=vtx=x0+vt=vt; Equation for the police car: This car is accelerating, so use the equation for displacement with x0=0x0=0 and v0=0v0=0, since the police car starts from rest: x=x0+v0t+12at2=12at2x=x0+v0t+12at2=12at2; Now we have an equation of motion for each car with a common parameter, which can be eliminated to find the solution. In this case, we solve for tt. Step 1, eliminating xx: x=vt=12at2x=vt=12at2; Step 2, solving for tt: t=2vat=2va. The speeding car has a constant velocity of 40 m/s, which is its average velocity. The acceleration of the police car is 4 m/s2. Evaluating t, the time for the police car to reach the speeding car, we have t=2va=2(40)4=20st=2va=2(40)4=20s.

97.

At this acceleration she comes to a full stop in t=v0a=80.5=16st=v0a=80.5=16s, but the distance covered is x=8m/s(16s)12(0.5)(16s)2=64mx=8m/s(16s)12(0.5)(16s)2=64m, which is less than the distance she is away from the finish line, so she never finishes the race.

99.

x1=32v0tx1=32v0t
x2=53x1x2=53x1

101.

v0=7.9m/sv0=7.9m/s velocity at the bottom of the window.
v=7.9m/sv=7.9m/s
v0=14.1m/sv0=14.1m/s

103.

a. v=5.42m/sv=5.42m/s;
b. v=4.64m/sv=4.64m/s;
c. a=2874.28m/s2a=2874.28m/s2;
d. (xx0)=5.11×103m(xx0)=5.11×103m

105.

Consider the players fall from rest at the height 1.0 m and 0.3 m.
0.9 s
0.5 s

107.

a. t=6.37st=6.37s taking the positive root;
b. v=59.5m/sv=59.5m/s

109.

a. y=4.9my=4.9m;
b. v=38.3m/sv=38.3m/s;
c. −33.3m−33.3m

111.

h=12gt2h=12gt2, h = total height and time to drop to ground
23h=12g(t1)223h=12g(t1)2 in t – 1 seconds it drops 2/3h
23(12gt2)=12g(t1)223(12gt2)=12g(t1)2 or t23=12(t1)2t23=12(t1)2
0=t26t+30=t26t+3 t=6±624·32=3±242t=6±624·32=3±242
t = 5.45 s and h = 145.5 m. Other root is less than 1 s. Check for t = 4.45 s h=12gt2=97.0h=12gt2=97.0 m =23(145.5)=23(145.5)

Challenge Problems

113.

a. v(t)=10t12t2m/s,a(t)=1024tm/s2v(t)=10t12t2m/s,a(t)=1024tm/s2;
b. v(2s)=−28m/s,a(2s)=−38m/s2v(2s)=−28m/s,a(2s)=−38m/s2; c. The slope of the position function is zero or the velocity is zero. There are two possible solutions: t = 0, which gives x = 0, or t = 10.0/12.0 = 0.83 s, which gives x = 1.16 m. The second answer is the correct choice; d. 0.83 s (e) 1.16 m

115.

96km/h=26.67m/s,a=26.67m/s4.0s=6.67m/s296km/h=26.67m/s,a=26.67m/s4.0s=6.67m/s2, 295.38 km/h = 82.05 m/s, t=12.3st=12.3s time to accelerate to maximum speed
x=504.55mx=504.55m distance covered during acceleration
7495.44m7495.44m at a constant speed
7495.44m82.05m/s=91.35s7495.44m82.05m/s=91.35s so total time is 91.35s+12.3s=103.65s91.35s+12.3s=103.65s.

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