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