Question

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 (magnitude and direction) of the ball during the 1.0 ms when it was in contact with the bat?

Answer 1

Assuming north as positive direction, the initial and final velocities of the ball are:

`v_i=-43 m/s` (with negative sign since it is due south)

`v_f=+51 m/s`
the time taken is
`t=1.0 ms=0.001 s`, so the average acceleration of the ball is given by

`a= (v_f-v_i)/(t)= (51 m/s-(-43 m/s))/(0.001 s)=9.4 \cdot 10^4 m/s^2`
And the positive sign tells us the direction of the acceleration is north.