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Magnitude of average acceleration

  1. Aug 23, 2006 #1
    A 50.7 g superball traveling at 25.1 m/s bounces off a brick wall and rebounds at 22.8 m/s. A high speed camera records this event. If the ball is in contact with the wall for 3.5 ms, what is the magnitude of the average acceleration of the ball during this time period?

    Unfortunately I'm stuck on another problem. For this one I first converted 3.5 ms to 0.0035s. I chose an arbitrary time, such as 4 seconds to give some perspective on time. I multiplied 25.1m/s x 4s to get a distance thinking that it took the ball 4s to hit the wall at that velocity. 25.1 x 4 = 100.4m. Then I determined the time it would take the ball to bounce back the same distance but at 22.8m/s using cross mulitplying and solving for t (time). 22.8m/ 1s = 100.4m/t. Then I used this info. to calculate average acceleration= (v2-v1)/ (t2-t1). I assume that when it asks for the magnitude it means absolute value, correct me if I'm wrong. The answer I got was 0.273 m/s^2 and it's wrong. I used this approach for another problem and it worked so let me know if this is at all the correct approach or if I'm completely off.

    Thanks
     
  2. jcsd
  3. Aug 23, 2006 #2
    I forgot to mention that I also included the 0.0035 in the total time to calculate the average acceleration assuming that initial time is 0s.
     
  4. Aug 23, 2006 #3

    Doc Al

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    Staff: Mentor

    This is all you need. What's v2 and v1? What's the elapsed time?

    The problem is much easier than you think. :wink:
     
  5. Aug 23, 2006 #4
    v2 and v1 are in opposite directions, right?
    So since v1 is a negative quantity, v2 -(-v1) = v2 + v1...
    Magnitude would just be the magnitude of the acceleration.. not the magnitudes of the velocities.
     
  6. Aug 23, 2006 #5
    The thing is I don't know the elapsed time. I took the 0.0035 s as when the ball is in contact with the wall.
     
  7. Aug 23, 2006 #6

    Doc Al

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    That is the elapsed time!
     
  8. Aug 23, 2006 #7
    You know I used that before and got it wrong then tried it again and it's right. I must be off on my math today because I was stressing over something I already knew how to do.
     
  9. Sep 9, 2009 #8
    I got a question similar to this and the way i did it was acceleration= velocity of the ball where its going (22.8 m/s) subtracted from the velocity before it hits the brick wall (25.1 m/s) divided by the time of 3.5ms. Please correct me if i am wrong.
     
  10. Sep 9, 2009 #9

    Doc Al

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    Assuming that you gave the two velocities different signs--since they are in opposite directions--your calculation should be correct.
     
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