What is the speed at which the belt of the ramp is moving?

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 ramp moves at a constant speed such that a person who stands still on it leaves the ramp 63.1 s after getting on. Clifford is in a real hurry, however, and skips the speed ramp. Starting from rest with an acceleration of 0.341 m/s2, he covers the same distance as the ramp does, but in one-fourth the time.

Answer 1

let's say the ramp overs a distance D in 63.1s with an average speed v then the equation expressing this motion is:

D= 63.1v

the equation describing Clifford's motion is

D=v0t+1/2at^2  where vo is the initial speed (here, 0), a is the accel (0.341m/s/s) and t is the time he traveled, given as 63.1/4=15.8s

so the distance covered is

D=1/2(0.341)(15.8)^2=42.4 m

using this above, we have that 42.4m=63.1v or that

v=0.67 m/s