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Direction and magnitude of the average acceleration

  1. Sep 25, 2011 #1
    1. A ball is dropeed from a height of 2.5m and rebounds to a height of 2.1m. If the ball is in contact with the floor for .70ms determine 1: direction and 2: the magnitude of the balls average acceleration due to the floor

    I know the equation for average acceleration = change in velocity/ time to make change

    I found initial velocity by: v(initial)= sqrt(2*9.8*2.5) = 7m/s
    I found final velocity by: v(final)= sqrt(2*9.8*2.1) = 6.42m/s

    When i plug those into the avg acceleration equation i have: (6.42m/s - 7 m/s)/.0007s
    to get: -828.57m/s So the direction would be down?

    Am i doing my math right I am not sure where the negatives go for gravity either. Thanks for your help.
     
  2. jcsd
  3. Sep 25, 2011 #2


    Okay, good start here. But you need to pick a sign convention. Usually we assume that positive is up. See what difference that makes.
     
  4. Sep 25, 2011 #3
    do you mean initial velocity should be: v(initial) = sqrt(2*-9.8m/s^2*-2.5m) = 7m/s
    and then final velocity would be the same, if so then am i right about the direction being down because the average acceleration is - ? Thanks for helping.
     
  5. Sep 25, 2011 #4

    lewando

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    Unfortunately, when you take the square root of a number you get a positive and negative result. This means you have to think about which of the two results applies to your system. Using Ignea_unda's recommendation, what would be the sign of the magnitude of the velocity upon initial contact with the ground?

    What would be the sign for the velocity as the ball stops contacting the ground?
     
  6. Sep 25, 2011 #5
    Upon initial contact with the ground the sign should be positive, i would think. When the ball hits the ground it is 0. As the ball moves upward, gravity is working on it again, but the ball is moving positively. What am I doing wrong here? It's just not clicking. I hope i'm close though thanks.
     
  7. Sep 25, 2011 #6

    lewando

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    The ball is moving downward-- it has downward velocity.

    This is true when the ball "bottoms out", somewhere in the .70ms interval.

    Yes, it is moving in the positive, up, direction. The velocity is pointing up.

    You are close. Just get the signs of the 2 velocities right.
     
  8. Sep 25, 2011 #7
    I found this on your forum:

    If velocity is (-) (down or to the left) and x (motion) is speeding up, then acceleration will have a (-) sign in front of it.
    If velocity is (+) (up or to the right) and x (motion) is slowing down, then acceleration will have a (-) sign in front of it.

    So my v(initial) should have negative sign in front and my v(final) should have a positive sign. correct? Then my avg acceleration = 6.42m/s-(-7m/s)/.0007 s giving a positive acceleration therefore a positive direction yes? I hope.
     
  9. Sep 25, 2011 #8

    lewando

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    Yep, good work by you!
     
  10. Sep 25, 2011 #9
    Yay! Thank you so much for your help! Have a great day!
     
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