Aspeed 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 64 s after getting on. clifford is in a real hurry, however, and skips the speed ramp. starting from rest with an acceleration of 0.37 m/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?

The belt ramp is moving at 0.047 m/s

Explanation:

Hi!

The equation for the position of an object moving in a straight line with a constant acceleration is:

x = x0 + v0 * t + 1/2 * a * t²

where:

x = position at time "t"

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

If the object moves with constant speed, then, a = 0 and x = x0 + v * t

First, let´s find the lenght of the speed ramp by calculating the distance walked by Clifford.

x = x0 + v0 * t +1/2 * a * t²

x0 = 0 placing the origin of our reference system at the begining of the ramp

v0 = 0 Clifford starts from rest

t = 64 s / 4  

a = 0.37 m/s²

Then:

x = 1/2 * 0.37 m/s² * 16 s = 3.0 m

Now that we know the lenght of the speed ramp, we can calculate the speed of the ramp which is constant:

x = x0 + v * t        x0 = 0

x = v * t

x/t = v

3.0 m / 64 s = 0.047 m/s

prolly a beacuse the momentuem the car has increases while going down hill

no. dot product   =0 when vecs at right angles