A baseball is pitched, reaching the batter with a speed of 37 m/s, and is hit such that it returns on the same line at 48 m/s. If the ball was in contact with the bat for 0.22 seconds, what was the magnitude of the average acceleration experienced by the ball during the time of contact? A) 220 m/s² B) 41 m/s² C) 390 m/s² D) 170 m/s²

Question

Quizplus · Accepted Answer

The average acceleration is calculated using the formula $a = \frac{\Delta v}{\Delta t}$, where $\Delta v$ is the change in velocity and $\Delta t$ is the change in time. The change in velocity is the final velocity minus the initial velocity. Since the ball changes direction, its final velocity (48 m/s towards the pitcher) is considered positive, and its initial velocity (37 m/s towards the batter) is considered negative, making the change in velocity $48 m/s - (-37 m/s) = 85 m/s$. The time of contact is 0.22 seconds. Thus, the average acceleration is $a = \frac{85 m/s}{0.22 s} = 386.36 m/s^2$, which rounds to 390 m/s^2, making the correct answer C.

A Baseball Is Pitched, Reaching the Batter with a Speed