Chapter 2: One-Dimensional Kinematics Selected Solutions

Selected Solutions

13. An arm is approximately 1 m in length so take Dx = 1 m.

29. (a) The runner starts from rest, and at t = 2.0 s the runner is still accelerating; thus

v = v₀ + at = 0 + (1.9 m/s²)(2.0 s) = 3.8 m/s

(b) The runner's speed at the end of the race is the same as that at t = 2.2 s.

v = v₀ + at = (1.9 m/s²)(2.2 s) = 4.2 m/s

35. (a) Since the car has constant deceleration, the equations for constant acceleration apply: Dt = Dv/a. Therefore, doubling the driving speed (Dv) will increase the time to stop by a factor of two.

(b)

The time required to stop is doubled.

85. (a) The youngster's time in the air is given by t_air = 2v₀/s. Therefore, if v₀ doubles so does t_air.

(b) The youngster's maximum height is given by . Therefore, if v₀ doubles then y_max increases by 2², so it quadruples.

(c) For part (a), t_air = 2v₀/g. Therefore,

For part (b), . Therefore,

91. (a) Just after release, the only acceleration is due to gravity. Therefore, a = 9.81 m/s², downward.

(b) At the maximum height, v = 0. Since we know v, v₀, and g, the appropriate expression to use is

Using this result, we have

(c) From the above result we can set y₀ = 15.5 m, y = 14 m, and v₀ = 0 then use the equations for free-fall to determine the time it takes for the shell to move between those two heights.

Since the shell goes both up to the maximum height and then back down to 14 m, we must double this result:

t_tot = 2t = 2(0.55 s) = 1.1 s

(d) Since, at a given level above the original height, the speed of an object in free-fall is the same coming down as it was going up, the speed when it returns to the 14 m height is 5.4 m/s.

Selected Solutions by David Reid, Eastern Michigan University. ©2002 by Prentice Hall, Inc.

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