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Math Help - Variable acceleration

  1. #1
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    Joined
    Sep 2008
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    239

    Variable acceleration

    Hi I just wondered if what i have done is correct. before i carry on doing these questions.

    1. A vehicle, moving along a straight horizontal road , has velocity v m/s at time t seconds modelled by

    v = \frac{1}{60}t(60-t) 0<=t<=60

    a) write down the times at which the vehicle is stationary

    b) calculate the distance travelled by the vehicle between t=0 and t=60

    c) i) Find, in terms of t, an expression for the acceleration of the vehicle at time t seconds

    ii) Find the greatest speed of the vehicle



    a) vehicle startionary at t=0 and t=60

    b) x=\int_0^{60} \frac{dx}{dt}

    \frac{1}{60}\int_0^{60} 60t-t^2dt

    = \frac{1}{60}(30t^2 - \frac{1}{3}t^3)_0^{60}

    = \frac{1}{60}(30(60^2)-\frac{1}{3}(60^3))

    =600m

    c) v = t - \frac{t^2}{60}

    \frac{dv}{dt} = 1 - \frac{2t}{60}

    greatest speed when a = 0

    1 - \frac{2t}{60} = 0

    60 - 2t = 0
    2t = 60
    t = 30s

    when t=30

    v = \frac{1}{60}(30)(60-30)

    = \frac{900}{60} = 15m.s
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  2. #2
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    Joined
    Apr 2009
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    a.) True when v(t) is 0, as you've shown at 0s and 60s

    b.) Integrate for s(t), initial distance is 0 so C = 0, answer is 600m

    c.) i.) v'(t) = a(t) ii.) occurs when a(t) = 0 and we go from positive to negative. True at t=30

    Looks good to me.
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