Question

A ball is rolling down a hill with an initial velocity of 9m/s. What is the acceleration of the ball if its displacement is 10m in 4s?

Answer 1

The displacement of a body moving with a constant acceleration is:

$$d = v_0t + \dfrac{at^2}{2}$$

where $v_0 = 9m/s$ is the initial velocity, $t = 4s$ is the time of movment and $a$ is the acceleration. Expressing $a$, obtain:

$$a = \dfrac{2d}{t^2} - \dfrac{2v_0}{t} = \dfrac{2\cdot 10m}{(4s)^2} - \dfrac{2\cdot 9m/s}{4s} = -3.25m/s^2$$

Answer. -3.25 m/s^2