Physics Question?

Question

A speed ramp at an airport is basically a large conveyor belt on which you can stand and be moved along. The belt of one speed ramp moves at a constant speed such that the person who stands still on it leaves the ramp 64 seconds after getting on. Clifford is on a real hurry, however, and skips the speed ramp. Starting from rest with an acceleration of 0.37 m/s^2, he covers the same distance as the ramp does, but does in one-fourth of the time. What is the speed at which the belt of the ramp is moving?

You may need the following equations:
v=v0+at
x=x0+v0t+1/2a(t^2)
v^2=v0^2+2ax

Answer 1

let t = time of Cliff's run. Then he covers the distance x in

x = 1/2at^2

Now the ramp takes 4t to cover the same distance so

x = vr*4t where vr is the speed of the ramp.

We are told that the ramp covers the distance x in 64 seconds thus 4t = 64 and t = 16 seconds.

So since Cliff and the ramp cover the same distance:

vr*4t = 1/2at^2 ---> vr =1/8at = 1/8*0.37 m/s^2*16 s = 0.74 m/s