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

Answer:

0.74 m/s

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

 $x = v_ot + \dfrac{1}{2}at^2$...(1)

initial velocity is equal to zero

 $x =\dfrac{1}{2}at^2$........(2)

equating equation (1) and (2)

$v_{belt}t_{belt}=\dfrac{1}{2}at^2$

$v_{belt}t_{belt}=\dfrac{1}{2}a(\dfrac{1}{4}\times t_{belt})^2$

$v_{belt}=\dfrac{1}{32}a t_{belt}$

$v_{belt}=\dfrac{1}{32}\times 0.37 \times 64$

$v_{belt}= 0.74\ m/s$

Speed of the belt is equal to 0.74 m/s