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Mechanics : Equations of motion.

  1. Apr 28, 2010 #1
    1. The problem statement, all variables and given/known data

    A motorcyclist starts from rest at a point O and travels in a straight line. His velocity after t seconds is vms^-2, for 0 =< t =< T, where v = 7.2t - 0.45t^2. The motorcyclist's accelaration is zero when t = T.

    Find the value of T.

    2. Relevant equations

    ---

    3. The attempt at a solution

    S = X
    U = 0
    V = 7.2t - 0.45t^2
    a = 0
    t = t

    v = u + at
    7.2t^2 - 0.45t = 0
    0.45t(t - 16) = 0
    Therefore t = 0 or 16

    But the answer is 8.. I believe I'm getting the accleration part incorrect but I just cannot spot it.
    Please help - thank you :)
     
    Last edited: Apr 29, 2010
  2. jcsd
  3. Apr 28, 2010 #2

    hotvette

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    Homework Helper

    The expression for v you used in step 3 doesn't match the problem statement. You swapped terms.
     
  4. Apr 29, 2010 #3
    Ahh apologies - I mistyped it. The second line is supposed to have the ^2 on the 0.45t :(
     
  5. Apr 29, 2010 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is only true for a constant acceleration- and a constant acceleration gives a linear velocity, not quadratic as you have here.

    All you have done here is solve for V= 0, not acceleration.

    The acceleration is the derivative of the velocity function. Take the derivative of [itex]7.2t^2- 0.45t[/itex] and set that equal to 0.
     
  6. May 2, 2010 #5
    Thank you for the reply. I remember my teacher had mention not to attempt that question as he had not covered it in class yet. I read up about it and you're help has reinforced my understanding.

    Thanks again. :)
     
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