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

Call the distance "d". This stands for the length of the conveyor belt, as well at the distance that Clifford travels.

And, let "v" be the speed of the conveyor, which is what we're trying to find.

Even though we don't know how much "d" and "v" are, we can certainly make this claim:

v = d/(64 sec)

Now, we are told that Clifford can cover the same distance in 1/4 the time (that is, 16 seconds). Using the formula for accelerated motion starting from rest:

Clifford's distance = (a)(16s)²/2

But Clifford's distance is "d" so:

d = (a)(16s)²/2

Combining this with the previous equation, we have:

v = d/(64 sec)
= ((a)(16s)²/2)/(64s)
= ((0.37m/s²)(16s)²/2)/(64s)