A rubber ball is attached to a paddle by a rubber band. The ball is initially moving away from the paddle with a speed of 5.5 m/s. After 0.35 s, the ball is moving toward the paddle with a speed of 5.5 m/s. What is the average acceleration of the ball during that 0.35 s? Give magnitude and direction.

Call away direction +

acceleration = change in velocity / change in time

= (Vfinal - Vinitial) / (Tfinal - Tinitial)

= (-5.5 - 5.5) / .35

A skateboard rider starts from rest and maintains a constant acceleration of 10.50 m/s2 for 8.4 s. what is the rider's displacement during this time?

To find the average acceleration of the ball during the given time interval, you will need to calculate the change in velocity and divide it by the time interval.

To calculate the change in velocity, subtract the initial velocity from the final velocity:

Change in velocity = Final velocity - Initial velocity

In this case, the initial velocity is +5.5 m/s (moving away) and the final velocity is -5.5 m/s (moving toward). Since they have the same magnitude but opposite directions, the change in velocity is:

Change in velocity = (-5.5 m/s) - (+5.5 m/s)
= -5.5 m/s - 5.5 m/s
= -11 m/s

Now, divide the change in velocity by the given time interval (0.35 s) to find the average acceleration:

Average acceleration = Change in velocity / Time interval

Average acceleration = (-11 m/s) / (0.35 s)
= -31.43 m/s^2

So, the average acceleration of the ball during the 0.35 s time interval is 31.43 m/s^2, directed toward the paddle.