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Homework Help: Kinematics of a point in rectilinear motion

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data
    A point Q in rectilinear motion passes through the origin at t=0, and from then until 5 seconds have passed, the acceleration of Q is 6 ft/s^2 to the right. Beginning at t = 5s, the acceleration of Q is 12t ft/s^2 to the left. If after 2 more seconds point Q is 13 feet to the right of the origin, what was the velocity of Q at t=0?

    2. Relevant equations

    3. The attempt at a solution

    At t=0:

    At t=5:

    At t=7:

    Now plug in t=5 s into the first set of equations and get:

    Plug these two equations into 13=-2t^3+(v1)(t)+x1
    As for t in this equation, plug in 2 seconds because this new equation started at t=5 (and hence assume that to be the new starting point, thus 7-5=2).

    I get v0 = -15 m/s
    The book says the answer is +2. Obviously something went very wrong. Any help would be appreciated.
  2. jcsd
  3. Mar 9, 2012 #2
    How did you get the highlighted equation ?
  4. Mar 9, 2012 #3
    plug t=5 into x'=6t+v0
    Then knowing the fact that at t=5 the velocity is described by x'=-6t^2+v1, I set t=0 at this instant and thus 30+v0 = v1
  5. Mar 9, 2012 #4
    you are using same constants on integration in different equations. the motion from t=5 to t=7
    should have separate constants of integration from the motion t=7 onwards. thats why you are
    mixing equations and got the wrong answer
  6. Mar 13, 2012 #5
    But according to the problem, the motion at t=7s follows the same equation as the motion at t=5s (x''=-12), so shouldn't it be the same equations? Can you show me the correct equation? Because from the way I'm reading the problem my way to me is correct, when in fact it's not. I can't see it.
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