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

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SUMMARY

The average acceleration of a 50.0 g Super Ball, which rebounds from a brick wall, is calculated using the formula for average acceleration: (final velocity - initial velocity) / change in time. Given the initial velocity of 30.0 m/s and a final velocity of 22.0 m/s, with a contact time of 5.00 ms, the calculation yields an average acceleration of 1600 m/s². The negative sign indicates direction, but the magnitude is the focus of the problem, confirming that the average acceleration is indeed 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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