We use the kinematic equation,
Here, s is the distance traveled by elevator u is the initial velocity and t is time.
As elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12 s, so
Thus, the distance the elevator travels during 12 s time is
Answer:
See the attachment below for the graphics in part (a)
The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.
This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.
Explanation:
The full solution can be found in the attachment below.
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