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Math Help - solving motion of type m x'' = F(t)

  1. #1
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    solving motion of type m x'' = F(t)

    A particle of mass 3 kg, initially at position y(0) = 2 and moving with velocity v(0) = 4) is subject to the exponential driving force F(t) = 4e^(-3t)
    Find an expression for y(t).

    I've got to the point where i've applied v(t) = V(0) + 1/M x (integral) f(t').
    no clue what to do after.
    any help would be appreciated


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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: solving motion of type m x'' = F(t)

    First start with F(t) = ma(t), or rearrange to get  a(t) = \frac {F(t)}{m} = \frac 4 3 e^{-3t} .

    Next  v = \int a(t) dt = \int \frac 4 3 e^{-3t} dt = - \frac 4 9 e^{-3t} + C. You can evalaute the constant of integration C from the initial condition that V(0) = 4
    m/s.

    Next:  y(t) = \int v(t)dt = \int (-\frac 4 9  e^{-3t} + C) dt = \frac 4 {27} e^{-3t} + C t + D and again you can find the value for the constant of integration D from the initial condition y(0) = 2 m.

    Hope this helps.
    Thanks from Unknownstarz
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