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 +(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 = (v_(f)- v_(i)  )/(t) = (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+(1)/(2) 0.208 \ m/s^2(12)^2 = 14.97 m \simeq  15 m`