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Average acceleration of the ball

  1. Jun 26, 2009 #1
    A 50.0 g Super Ball traveling at 25.0 m/s bounces off a brick wall and rebounds at 21.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 5.00 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms = 10-3 s.)

    a=change in velocity/change in time


    What I did so far was (21-25)/0.005=-800 m/s^2
    however this wrong, not sure why though. Could someone explain why my attempt is wrong
     
  2. jcsd
  3. Jun 26, 2009 #2

    Matterwave

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    Re: Acceleration

    When considering velocity, direction matters as well. When the ball is traveling TO the brick wall, it is going in another direction than when it is traveling FROM the brick wall. How would you account for this change in direction?
     
  4. Jun 26, 2009 #3

    jgens

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    Re: Acceleration

    You're calculating the change in velocity wrong - you need to be careful with your signs. To pick a convention, assume that the ball is initially traveling in the positive direction, this way its initial velocity Vi = 25 m/s. When the ball rebounds, it travels in the opposite or negative direction so we say the final velocity Vf = 21 m/s.

    By definition we have that ΔV = Vf - Vi. Use this value to calculate the average acceleration.
     
  5. Jun 26, 2009 #4
    Re: Acceleration

    So would one of the velocities be negative. So would it be +25 and -21, and t=0.005?
     
  6. Jun 26, 2009 #5
    Re: Acceleration

    so would Aave=-21-25/0.005=-9200
     
  7. Jun 26, 2009 #6
    Re: Acceleration

    -9200m/s^2
     
  8. Jun 26, 2009 #7
    Re: Acceleration

    however this is not right? what am I doing wrong?
     
  9. Jun 26, 2009 #8

    jgens

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    Re: Acceleration

    Since the question asks for the magnitude of the average acceleration (direction doesn't matter) you don't need the negative sign.
     
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