Motion and Average Acceleration -Help-

[febreze]

I've been killing myself over this problem because I cannot figure it out. Can someone please help me?


A tennis ball with a velocity of 40.0 m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball rebounds in the opposite direction with a velocity of -8 m/s to the left. If the ball is in contact with the wall for 0.014 s, what is the average acceleration of the ball while it is in contact with the wall?

 

[zwtipp05]

Have you learned about momentum and impulse?

Force multipled by time gives you change in momentum.
$$F\Delta{t}=mv_f-mv_i$$

Then just use $$F=ma$$

 

[quicknote]

Hi there, I can understand your frustration with this problem. Let me try to help you out.

First, it is important to understand that acceleration is the rate of change of velocity over time. In this case, the ball's initial velocity is 40.0 m/s to the right, and its final velocity is -8 m/s to the left. This means that the ball's velocity changed by 48 m/s (40 m/s - (-8 m/s)) in the opposite direction during its contact with the wall.

Next, we need to determine the time it took for this change in velocity to occur. We are given that the ball was in contact with the wall for 0.014 s. Therefore, the average acceleration can be calculated by dividing the change in velocity by the time taken: 48 m/s / 0.014 s = 3428.57 m/s^2.

So, the average acceleration of the ball while in contact with the wall is 3428.57 m/s^2 in the direction opposite to its initial velocity. I hope this helps you solve the problem. Keep practicing and don't give up!