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 constants speed such that a person who stands still on it leaves the ramp 68.5s after getting on. Clifford is in a real hurry, however, and skips the speed ramp. Starting from rest with an acceleration of 0.333m/s2, he covers the same distance as the ramp does, but in one-fourth the time. What is the speed at which the belt of the ramp is moving?
Let's denote the speed of the ramp's belt as v (m/s). The time Clifford takes to cover the same distance as the ramp does is 1/4 of the time it takes for a person standing still on the ramp.
Given that the time for a person standing still is 68.5 seconds, Clifford's time is 1/4 of 68.5 seconds = 68.5/4 = 17.125 seconds.
Now, we can use the equations of motion to find the distance covered by Clifford. We know that an object's displacement (d) is given by:
d = ut + (1/2)at^2,
where u is the initial velocity, a is the acceleration, and t is the time.
Since Clifford starts from rest, his initial velocity (u) is 0 m/s, and his acceleration (a) is 0.333 m/s^2.
Substituting the values, we have:
d = 0(17.125) + (1/2)(0.333)(17.125)^2
d = (1/2)(0.333)(293.265)
d = 48.718 m.
Now, let's find the speed (v) at which the ramp's belt is moving using the equation:
v = d/t,
where d is the distance and t is the time.
Substituting the values:
v = 48.718/68.5
v ≈ 0.711 m/s.
Therefore, the speed of the ramp's belt is approximately 0.711 m/s.