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Motion Equation Problem?

  1. Sep 16, 2009 #1
    How would you solve a problem such as this?

    A Honda and a Porsche race, starting from the same point. The Honda accelerates at a constant 4.00 m/s2; the Porsche at a constant 8.00 m/s2. The Porsche gives the Honda an advantage by letting it start first. The Honda accelerates, and when it is traveling at 23.0 m/s, the Porsche starts. How far do the cars travel from the starting point before the Porsche catches up with the Honda?

    All I have is this:

    honda v = 4x = 23 m/s
    x = 5.75 s

    honda position = 2x^2
    honda (5.75) = 2(5.75)^2 = 66.125 m

    so all I know is when the honda is 66.125 meters the porche starts to go.

    what do I next?
     
  2. jcsd
  3. Sep 16, 2009 #2

    ideasrule

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    Homework Helper

    Now, you imagine that you are in the Porsche. You see yourself accelerating towards the Honda at 8-4=4 m/s^2 and have to travel 66.125 m before catching up to it.
     
  4. Sep 16, 2009 #3

    rl.bhat

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    When they meet after time t,
    Honda travels a distance
    d = vo*t + 0.5*a1*t^2 Here vo = 23 m/s .........(1)
    Porsche travels a distance
    d + 66.125 = vo*t + 0.5*a2*t^2. Here vo = 0......(2)
    From eq.1 and 2, find t.
     
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