A 50-g ball traveling at 24.0 m/s is bounced off a brick wall and rebounds at 19.0 m/s.A high-speed camera records this event.If the ball is in contact with the wall for 4.00 ms,what is the average acceleration of the ball during this time interval?

Question

Quizplus · Accepted Answer

The average acceleration of the ball is given by the change in velocity divided by the time interval:$$a = \frac{\Delta v}{\Delta t} = \frac{19.0 m/s - 24.0 m/s}{4.00 ms} = -10,750 m/s^2$$The negative sign indicates that the acceleration is in the opposite direction of the ball's initial velocity.

A 50-G Ball Traveling at 24