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Distance after pushoff between two objects

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data

    A 60 kg parent and a 12 kg child meet at the center of an ice rink. They push off each other for .5 seconds. If the acceleration of the child is 2.6 m/s^2, then what is the distance between the parent and the child at 3 seconds after their push off?


    2. Relevant equations

    F=ma
    X=Xo +Vot + 1/2at^2



    3. The attempt at a solution

    I know that the net force of the child is (12)(2.6)= 31.2 N. Using Newtons equal and opposite law, I know that the force of the parent must be -31.2 N. Then to find the acceleration of the parent, -31.2=(60)a which means a= -.52 m/s^2. From there I think you plug in each of of the acceleration with time 2.5 into the equation above to get each of their positions (X=0 + 0 + 1/2 at^2). with that I get about 6 meters difference and I know that is not the answers
     
  2. jcsd
  3. Feb 24, 2010 #2

    tiny-tim

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    Hi spider3367! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Yes, you've basically done it correctly, but you've not read the question properly! :rolleyes:

    The acceleration is as given for .5 seconds, but after that, the acceleration is zero. :wink:

    (also, you could have got .52 m/s2 simply by using good ol' Newton's first law … there's no external force, so the position of the center of mass will be stationary)

    Try again! :smile:
     
  4. Feb 24, 2010 #3
    So I'm still a little confused.

    (2.6 m/s2)(.5^2)*1/2=.325
    (.52m/s2)(.5^2)*1/2=.065

    add them= .39 meters? That seems like it can't be right.
     
  5. Feb 24, 2010 #4

    tiny-tim

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    (please use the X2 tag just above the Reply box :wink:)

    i] you could have added the 2.6 to the .52 to get the relative acceleration, abnd put that into the formula

    ii] you still have another 2.5 seconds of constant speed to account for. :smile:

    (and i'm going to bed :zzz:)​
     
  6. Feb 24, 2010 #5
    Am I on the right track with this:

    xtotal=[0+(2.6)(2.5)] + [0+(.52)(2.5)] + .39 meters (from calculated accelerations)
     
  7. Feb 25, 2010 #6

    tiny-tim

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    Hi spider3367! :smile:

    (just got up :zzz: …)
    Nooo … you need the speed after 0.5 seconds.
     
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