Probleas de física

2
Motion in One Dimension
CHAPTER OUTLINE
2.1 2.2 2.3 2.4 2.5 2.6 2.7 Position, Velocity, and Speed Instantaneous Velocity and Speed Acceleration Motion Diagrams One-Dimensional Motion with Constant Acceleration Freely Falling Objects Kinematic Equations Derived from Calculus

ANSWERS TO QUESTIONS
Q2.1 If I count 5.0 s between lightning and thunder, the sound has traveled 331 m s 5.0 s =1.7 km . The transit time for the light is smaller by

b

ga f

3.00 × 10 8 m s = 9.06 × 10 5 times, 331 m s so it is negligible in comparison. Q2.2 Q2.3 Q2.4 Yes. Yes, if the particle winds up in the +x region at the end. Zero. Yes. Yes.

Q2.5

No. Consider a sprinter running a straight-line race. His average velocity would simply be the length of the race divided by the time it took forhim to complete the race. If he stops along the way to tie his shoe, then his instantaneous velocity at that point would be zero. We assume the object moves along a straight line. If its average x velocity is zero, then the displacement must be zero over the time interval, according to Equation 2.2. The object might be stationary throughout the interval. If it is moving to the right at first, itmust later move to the left to return to its starting point. Its velocity must be zero as it turns around. The graph of the motion shown to the right represents such motion, as the initial and final positions are the same. In an x vs. t graph, the instantaneous velocity at any time t is the slope of the curve at that point. At t 0 in the graph, the slope of the curve is zero, and thus theinstantaneous velocity at that time is also zero.

Q2.6

t0

t

FIG. Q2.6 Q2.7 Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, or is a constant.

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22 Q2.8

Motion in One Dimension

Yes. If you drop a doughnut from rest v = 0 , then its acceleration is not zero. A common misconception isthat immediately after the doughnut is released, both the velocity and acceleration are zero. If the acceleration were zero, then the velocity would not change, leaving the doughnut floating at rest in mid-air. No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedalin the recent past. Yes. Consider throwing a ball straight up. As the ball goes up, its v velocity is upward v > 0 , and its acceleration is directed down v0 a < 0 . A graph of v vs. t for this situation would look like the figure to the right. The acceleration is the slope of a v vs. t graph, and is always negative in this case, even when the velocity is positive.

a

f

Q2.9

Q2.10

a fa

f

t FIG. Q2.10 Q2.11 (a) (d) (g) (h) Q2.12 Q2.13 Accelerating East Braking West (b) (e) Braking East Accelerating West (c) (f) Cruising East Cruising West

Stopped but starting to move East Stopped but starting to move West

No. Constant acceleration only. Yes. Zero is a constant. The position does depend on the origin of the coordinate system. Assume that the cliff is 20 m tall,and that the stone reaches a maximum height of 10 m above the top of the cliff. If the origin is taken as the top of the cliff, then the maximum height reached by the stone would be 10 m. If the origin is taken as the bottom of the cliff, then the maximum height would be 30 m. The velocity is independent of the origin. Since the change in position is used to calculate the instantaneous velocity inEquation 2.5, the choice of origin is arbitrary. Once the objects leave the hand, both are in free fall, and both experience the same downward acceleration equal to the free-fall acceleration, –g. They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity equal to vi . This velocity is the same as the velocity of the second...