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2.1: a) During the later 4.75-s interval, the rocket moves a distance
[pic], and so the magnitude of the average velocity is
[pic]
b) [pic]

2.2: a) The magnitude of the average velocity on the return flight is

[pic]

The direction has been defined to be the –x-direction [pic]

b) Because the bird ends up at the starting point, the average velocityfor the round
trip is 0.

2.3: Although the distance could be found, the intermediate calculation can be avoided by considering that the time will be inversely proportional to the speed, and the extra time will be
[pic]
2.4: The eastward run takes [pic] = 40.0 s and the westward run takes [pic] = 70.0 s. a) (200 m + 280 m)/(40.0 s + 70.0 s) = [pic] to two significantfigures. b) The net displacement is 80 m west, so the average velocity is [pic] = [pic] in the –x-direction [pic]

2.5: In time t the fast runner has traveled 200 m farther than the slow runner:
[pic].
Fast runner has run [pic]
Slow runner has run [pic]

2.6: The s-waves travel slower, so they arrive 33 s after the p-waves.

[pic]

2.7: a) The van will travel 480 m for the first 60 s and1200 m for the next 60 s, for a total distance of 1680 m in 120 s and an average speed of [pic] b) The first stage of the journey takes [pic] and the second stage of the journey takes [pic] so the time for the 480-m trip is 42 s, for an average speed of [pic] c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same.

2.8: From theexpression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) [pic] b) [pic] c) [pic]

2.9: a) At [pic], so Eq (2.2) gives
[pic]

b) From Eq. (2.3), the instantaneous velocity as a function of time is

[pic]

so i) [pic]
ii) [pic]
and iii) [pic]
c) The car is at rest when [pic]. Therefore [pic]. The only time after [pic] when the car isat rest is [pic]

2.10: a) IV: The curve is horizontal; this corresponds to the time when she stops. b) I: This is the time when the curve is most nearly straight and tilted upward (indicating postive velocity). c) V: Here the curve is plainly straight, tilted downward (negative velocity). d) II: The curve has a postive slope that is increasing. e) III: The curve is still tilted upward(positive slope and positive velocity), but becoming less so.

2.11: Time (s) 0 2 4 6 8 10 12 14 16
Acceleration (m/s2) 0 1 2 2 3 1.5 1.5 0

a) The acceleration is not constant, but is approximately constant between the times [pic] and [pic]

b)

[pic]

At [pic]

2.12: The cruising speed of the car is 60 [pic] = 16.7[pic]. a) [pic] (to two significant figures). b) [pic]c) No change in speed, so the acceleration is zero. d) The final speed is the same as the initial speed, so the average acceleration is zero.

2.13: a) The plot of the velocity seems to be the most curved upward near t = 5 s.
b) The only negative acceleration (downward-sloping part of the plot) is between t = 30 s and t = 40 s. c) At t = 20 s,the plot is level, and in Exercise 2.12 the car is said to be cruising at constant speed, and so the acceleration is zero. d) The plot is very nearly a straight line, and the acceleration is that found in part (b) of Exercise 2.12, [pic]
e)
[pic]

2.14: (a) The displacement vector is:

[pic]

The velocity vector is the time derivative of the displacement vector:

[pic]

andthe acceleration vector is the time derivative of the velocity vector:

[pic]

At t = 5.0 s:

[pic]
[pic]
[pic]
[pic]

(b) The velocity in both the x- and the y-directions is constant and nonzero; thus the overall velocity can never be zero.
(c) The object's acceleration is constant, since t does not appear in the acceleration vector....
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