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4 years ago

9

A softball player is running at 4.88 m/sec when she slides into second base coming to a stop in .872 seconds. How far did she sl

ide, and what was her acceleration?

1 answer:

kati45 [8]4 years ago

6 0

Answer:

d=v1t - .5at^2

d=4.88 x .872 - 0.5 x (4.88/0.872) x 0.872^2

d=4.255 - 2.12

d= 2.135m

Explanation:

acceleration is negative because she is slowing down.

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We need to calculate the acceleration at point A

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Put the value in the equation

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$a_{A}=-0.146j-0.215i−0.636i+0.937j$

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$a_{A}=-0.851i+0.791j\ m/s^2$

Hence, (a). The velocity at point A is $(-0.267j-0.393i)\ m/s$

(b). The acceleration at point A is $(-0.851i+0.791j)\ m/s^2$

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