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## Homework Statement

A 50.0-g superball is travelling at 25.0 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 3.50 ms (.0035 s) what is the magnitude of the average acceleration of the ball during this time interval?

## Homework Equations

[tex] \bar{a}= \frac {\Delta v}{\Delta t} = \frac {v_{f}-v_{i}}{t_{f}-t_{i}} [/tex]

## The Attempt at a Solution

So the information from the problem gives me:

[tex] v_{i}=25 \frac{m}{s} [/tex]

[tex] v_{f} = 22 \frac {m}{s} [/tex]

[tex] t_{i} = 0 s [/tex]

[tex] t_{f} = .0035s [/tex]

so the average acceleration during the time interval should be:

[tex] \bar{a} = \frac {22-25}{.0035-0}=|-\frac {3}{.0035}| \approx 857.14 \frac {m}{s^{2}} [/tex]

only problem is that the book has 1.34 x 10

^{4}m/s^2 as the answer, what am I missing here?

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