Question

004 10.0 points
a tennis ball with a speed of 20.6 m/s is moving perpendicular to a wall. after striking the wall, the ball rebounds in the opposite direction with a speed of 11.948 m/s.
if the ball is in contact with the wall for 0.0082 s, what is the average acceleration of the ball while it is in contact with the wall? take \toward the wall\ to be the positive direction.
answer in units of m/s².

Explanation:

Step1: Define initial and final velocities

Let the initial velocity $v_0 = 20.6\ m/s$ (toward the wall, positive), and the final velocity $v = - 11.948\ m/s$ (away from the wall, negative).

Step2: Use acceleration formula

The formula for average acceleration is $a=\frac{v - v_0}{t}$, where $t = 0.0082\ s$. Substitute the values: $a=\frac{-11.948 - 20.6}{0.0082}$.

Step3: Calculate the acceleration

First, calculate the numerator: $-11.948-20.6=-32.548$. Then, divide by the time: $a=\frac{-32.548}{0.0082}=-3969.268\ m/s^{2}$. The negative sign indicates the direction is away from the wall.

Answer:

$-3969.27\ m/s^{2}$