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Velocity and Time

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Boxes are placed on a chute at a uniform rate of time Tr and slide down the chute with uniform acceleration. Knowing that as any box B is released, the preceeding box A is 6m down the chute; 1 second later there is a 10m distance between them. Determine the value of Tr

    2. Relevant equations

    3. The attempt at a solution

    As x = 0.5at2
    6 = 0.5a(Tr)2

    I found a to be (change in y) / (change in x) = 4 m/sec2

    So, [tex]\sqrt{6/(0.5*0.4)}[/tex] = 3^(0.5) = Tr

    But the answer is plain old 3...

    Any advice appreciated :)
  2. jcsd
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