Question

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 cov- ers 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?

Answer 1

0.74 m/s

Step-by-step explanation:

given,

time taken by the ramp to cover cliff , t = 64 s

acceleration = 0.37 m/s²

distance traveled by the belt

`x = v_(belt)t_(belt)`

Clifford is moving with constant acceleration

`x = v_ot + (1)/(2)at^2`...(1)

initial velocity is equal to zero

`x =(1)/(2)at^2`........(2)

equating equation (1) and (2)

`v_(belt)t_(belt)=(1)/(2)at^2`

`v_(belt)t_(belt)=(1)/(2)a((1)/(4)* t_(belt))^2`

`v_(belt)=(1)/(32)a t_(belt)`

`v_(belt)=(1)/(32)* 0.37 * 64`

`v_(belt)= 0.74\ m/s`

Speed of the belt is equal to 0.74 m/s