Magnitude of Average Acceleration

[jets29]

A 50.0 g Super Ball traveling at 27.0 m/s bounces off a brick wall and rebounds at 15.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.50 ms, what is the magnitude of the average acceleration of the ball during this time interval?

I can't seem to get this problem right. Initially i converted 4.5ms to .0045s. Then I chose a time such as 4s to determine a distance and have a time perspective. I then multiplied 27x4 to determine the distance and got 108m, Then I determined the time it would take for the ball to return and got 7.2s. Then I used all this info with the .0045 included in the total time to calculate the average acceleration= (vf-vi)/(tf-ti). when everything was plugged in, (15-27)/(11.2045-0)=1.071m/s^2(assuming the absolute value is what they want when they say magnitude)
This is wrong. Any help will do thanks.

 

[jets29]

never mind i got it

 

[baba89]

Your approach is mostly correct, but there are a few errors in your calculations. Here's a step-by-step solution:

1. First, convert 4.50 ms to seconds: 4.50 ms = 0.00450 s.

2. Next, calculate the distance traveled by the ball during the 4.50 ms of contact with the wall. This can be done using the formula d = v_avg * t, where d is the distance, v_avg is the average velocity, and t is the time. In this case, the average velocity is (27.0 + 15.0)/2 = 21.0 m/s. Therefore, the distance traveled by the ball during the 4.50 ms of contact with the wall is:
d = 21.0 m/s * 0.00450 s = 0.0945 m.

3. Now, we need to determine the time interval for which the ball is in motion. This is the total time, which includes the 4.50 ms of contact with the wall, as well as the time it takes for the ball to travel from the wall back to its initial position. This can be calculated as follows:
Total time = 4.50 ms + (distance traveled by the ball / average velocity)
= 0.00450 s + (0.0945 m / 21.0 m/s)
= 0.00450 s + 0.00450 s
= 0.00900 s.

4. Now, we can use the formula for average acceleration, a_avg = (vf - vi)/t, where a_avg is the average acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval. In this case, vf = 15.0 m/s, vi = 27.0 m/s, and t = 0.00900 s. Therefore, the average acceleration of the ball during the 4.50 ms of contact with the wall is:
a_avg = (15.0 m/s - 27.0 m/s)/0.00900 s
= -12.0 m/s^2.

5. Since the question asks for the magnitude of the average acceleration, we need to take the absolute value of the result we obtained in step 4. Therefore, the magnitude of the average acceleration of the ball during the 4.50 ms of contact with