A pitcher delivers a fast ball with a velocity of 43 m/s to the south. The batter hits the ball and gives it a velocity of 51 m/s to the north. What was the average acceleration of the ball during the 1.0 ms when it was in contact with the bat? A)4.3 × 10⁴m/s², south B)5.1 × 10⁴m/s², north C)9.4 × 10⁴m/s², north D)2.2 × 10³m/s², south E)7.0 × 10³m/s², north

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Quizplus · Accepted Answer

The change in velocity is 51 m/s (north) - (-43 m/s (south)) = 94 m/s (north). The time is 1.0 ms = 0.001 s. Average acceleration = change in velocity / time = 94 m/s / 0.001 s = 9.4 × 10^4 m/s², north.

A Pitcher Delivers a Fast Ball with a Velocity of 43