2. An elevator is moving upwards at a constant speed of 6.0 m/s when a bolt come

Question

2. An elevator is moving upwards at a constant speed of 6.0 m/s when a bolt comes loose from the bottom. The bolt hits the bottom of the elevator shaft 3.0 s later. The elevator continues to move upward at constant speed during those three seconds.
a) Sketch graphs of the position, velocity and acceleration of the bolt, and of the bottom of the elevator, from the time the bolt came loose until the time it hit the bottom of the shaft.

b)  How high was the bolt above the bottom of the shaft when it came loose?

c)  What was the distance between the bolt and the bottom of the elevator when the

bolt reached its highest point?

Explanation / Answer

The initial velocity is 6 m/s (because the bolt was moving up with the elevator), the time is 3 seconds, and the acceleration is the gravitational acceleration, which is 9.81 m/s^2. Plug those into the formula, and you should find that the bolt fell about 26 m.

d = v-initial(t) + (a)(t^2)/2
d = 6(3) + (-9.81)(9)/2
d = 18 - 44.145
d = 26.145 m [down]

(b) This is asking for the final velocity. The formula is v-final^2 = v-initial^2 + 2(a)(d).

Again, your initial velocity is 6 m/s, your acceleration is 9.81 m/s^2, and your change in position is 26.145 m, as determined from part (a).

Plug those in, and you should get about 23 m/s.

v-final^2 = v-initial^2 + 2(a)(d)
v-final^2 = 36 + 512.9649
v-final = 23.43 m/s

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