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Math Help - Kinematics

  1. #1
    Member helloying's Avatar
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    Kinematics

    A particle travelling in a straight line passes a fixed point O with a velocity of 8m/s and a accerlation of -2m/ s^2. It moves in such a manner that t seconds after passing O its accerlation, fm/s^2 , is given by f= a + bt . During the 1st second after passing O it moves through 9m. calculate the value of a and of b.
    Last edited by helloying; June 4th 2009 at 02:39 AM.
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  2. #2
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    Quote Originally Posted by helloying View Post
    A particle travelling in a straight line passes a fixed point O with a velocity of 8m/s and a accerlation of -2m/ s^2. It moves in such a manner that t seconds after passing O its accerlation, fm/s^2 , is given by f= a + bt . Mr F says: I like using a for acceleration so I'm gonna change this to {\color{red}\alpha + \beta t}.

    During the 1st second after passing O it moves through 9m. calculate the value of a and of b. I dont really understand this question. This qn is so long and i cant imagine how the particle move.
    Let t = 0 be when the particle passes the fixed point O. Then you have:

    a = \alpha + \beta t subject to the boundary conditions that when t = 0 \, \text{s}, x = 0 \, \text{m}, v = 8 \, \text{m/s} and a = -2 \, \text{m/s}^2.

    t = 0 \, \text{s}: a = \alpha \Rightarrow \alpha = -2. Therefore a = -2 + \beta t.

    a = \frac{dv}{dt} = -2 + \beta t \Rightarrow v = -2t + \frac{\beta}{2} t^2 + C.

    Substitute v = 8 \, \text{m/s} when t = 0 \, \text{s}: C = 8. Therefore v = -2t + \frac{\beta}{2} t^2 + 8.

    v = \frac{dx}{dt} = -2t + \frac{\beta}{2} t^2 + 8 \Rightarrow x = -t^2 + \frac{\beta}{6} t^3 + 8t + K.

    Substitute x = 0 \, \text{m} when t = 0 \, \text{s}: K = 0.

    Therefore x = -t^2 + \frac{\beta}{6} t^3 + 8t.

    Now substitute x = 9 when t = 1 and solve for \beta.
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