Question

A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?

Answer 1

Answer:

  Ball will move 92.8125 meter along the cliff in 7.5 seconds.

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case initial velocity = 0 m/s, acceleration = 3.3 $m/s^2$, we need to calculate displacement when time = 7.5 seconds.

Substituting

  $s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter$

  So ball will move 92.8125 meter along the cliff in 7.5 seconds.