Question

An elevator accelerates uniformly from rest to a speed of 2.5 m/s in 12s. What is the distance the elevator travels during this time? must show work.

Answer 1

We use the kinematic equation,

$s= u t +\frac{1}{2}a t^2$                      

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

$a = \frac{v_{f}- v_{i} }{t} = \frac{2.5 m/s-0}{12 s} = 0.208 \ m/s^2$

Thus, the distance the elevator travels during 12 s time is

$s= 0+\frac{1}{2} 0.208 \ m/s^2(12)^2 = 14.97 m \simeq 15 m$