What Is the Average Acceleration of a Super Ball Bouncing Off a Wall?

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The discussion centers on calculating the average acceleration of a Super Ball that rebounds off a wall. The ball, initially traveling at 30.0 m/s and rebounding at 22.0 m/s, is in contact with the wall for 5.00 ms. The average acceleration is calculated using the formula (final velocity - initial velocity) divided by the time interval, resulting in -1600 m/s². It is emphasized that attention to the direction of velocity is crucial, as velocity is a vector quantity. The final answer for the magnitude of the average acceleration is 1600 m/s².
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Homework Statement



A 50.0 g Super Ball traveling at 30.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 5.00 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms = 10-3 s.)


Homework Equations



avg acceleration=(final V-initial V)/change in time

The Attempt at a Solution



(22m/s-30m/s)/0.005s= -1600m/s^2 and since it asks for the magnitude my final answer is
1600m/s^2

This was from my online physics homework. I really don't see what I'm doing wrong.
 
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Rob21 said:

Homework Statement



A 50.0 g Super Ball traveling at 30.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 5.00 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms = 10-3 s.)

Homework Equations



avg acceleration=(final V-initial V)/change in time

The Attempt at a Solution



(22m/s-30m/s)/0.005s= -1600m/s^2 and since it asks for the magnitude my final answer is
1600m/s^2

This was from my online physics homework. I really don't see what I'm doing wrong.

I think you need to look at your direction signs more carefully.

Remember velocity is a vector. The sign you assign it needs to be carried through the calculation.
 
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