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3 years ago

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A football player carrying the ball runs straight ahead at the line of scrimmage and directly into a wall of defensive linemen.

The ball carrier has an initial speed of 7.36 m/s and is stopped in a time interval of 0.180 s. Find the magnitude and direction of his average acceleration.

1 answer:

Colt1911 [192]3 years ago

4 0

solution:

$we have the following formula for average acceleration\\a_{ave}=\frac{v_{f}-v_{i}}{t}\\=\frac{0-(7.36)}{0.180}\\=-40.88m/s^2\\$

Now the direction of this signed acceleration is towards in the direction of travel of the ball carrier.

But as this acceleration is negative we could just as validly state that the acceleration is $40.88m/s^2$ in the opposite direction to the ball carrier's travel.

Although both are valid, the latter statement is probably the answer you are expected to give .

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<u>Using the conservation of linear momentum:</u>

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<u /> $KE_A=\frac{1}{2}m_A.v_A^2$ <u />

<u /> $KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}$ <u />

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