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Math Help - another linear motion question

  1. #1
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    another linear motion question

    A particle moves vertically under gravity and a retarding force proportional to the square of the velocity. If v is its upward or downward speed and we know that \dot{v}=-g-kv^2, show that its position at time t is given by
    z = z_0 + \frac{1}{k}ln\cos[\sqrtgk (t_0 -t)]
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    What have you done so far?
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  3. #3
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    I integrated with respect with time twice
    \frac{dv}{dt}=-g-kv^2
    v= v_0 - gt -kv^2t
    x= x_0 + v_0t -\frac{gt^2}{2} -\frac{-kv^2t^2}{2}+

    however i feel there is a technique to solce DEs that i dont know, because i cannot see where the ln cos... term would come from
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  4. #4
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    Your technique for solving the DE is, alas, entirely wrong. The reason is that the function v occurs in the RHS. You can't just integrate with respect to t, and treat v like a constant, when it varies with t. This equation is separable. Try

    \displaystyle{\int\frac{dv}{-g-kv^{2}}=\int dt}.
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  5. #5
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    I integrate the above and get
    \frac{-1}{k}\int \frac{1}{v^2 + \frac{g}{k}} dv = \int dt
    \frac{-1}{\sqrt (gk)}arctan (\frac{\sqrt (k)}{\sqrt(g)}v) = t - t_0
    arctan(\frac{\sqrt (k)}{\sqrt (g)}v)= \sqrt (gk) (t_0 -t)

    however from here taking the tangent of both sides does not leave anything readily integrable
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  6. #6
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    however from here taking the tangent of both sides does not leave anything readily integrable
    I think it does. You basically have the tangent of a linear function of t. So the numerator looks like the derivative of the denominator. What does that give you?
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  7. #7
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    thanks
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  8. #8
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    You're welcome. Have a good one!
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