Question

004 10.0 points a tennis ball with a speed of 19.3 m/s is moving perpendicular to a wall. after striking the wall, the ball rebounds in the opposite direction with a speed of 13.6258 m/s. if the ball is in contact with the wall for 0.0111 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: Recall the acceleration formula

Acceleration \( a \) is defined as the change in velocity \( \Delta v \) over the change in time \( \Delta t \), so \( a=\frac{\Delta v}{\Delta t}=\frac{v_f - v_i}{\Delta t} \). Here, the initial velocity \( v_i \) (toward the wall, positive direction) is \( 19.3\ m/s \), the final velocity \( v_f \) (away from the wall, so negative) is \( - 13.6258\ m/s \), and \( \Delta t = 0.0111\ s \).

Step2: Calculate the change in velocity

\( \Delta v=v_f - v_i=- 13.6258 - 19.3=-32.9258\ m/s \)

Step3: Calculate the acceleration

Substitute \( \Delta v \) and \( \Delta t \) into the acceleration formula: \( a = \frac{-32.9258}{0.0111}\approx - 2966.29\ m/s^{2} \) (the negative sign indicates the acceleration is in the direction opposite to the initial velocity, i.e., away from the wall)

Answer:

\( -2966.29\) (or approximately \(-2.97\times10^{3}\) depending on significant figures, but following the calculation above, the value is approximately \(-2966.29\ m/s^{2}\))