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Homework Help: Another problem from Kline

  1. Feb 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A body has acquired a downward velocity [itex] v_0 [/itex] and thereafter the acceleration is proportional to the distance traveled. Measure time and distance from the instant and place where the object has acquired the velocity [itex] v_0 [/itex] and find the formula or the distance traveled as a function of time.


    2. Relevant equations
    Suggestion in text to use

    [tex] \frac{dv}{dt} = v\, \frac{dv}{dy} [/tex]


    3. The attempt at a solution
    I have found this problem very challenging. I'd appreciate any suggestions. I know it's long and complicated, and I am asking a lot, but if any of you have time, it really would be appreciated.

    Using the suggestion in the text, I started with [itex] a = ky [/itex] which means [itex] \frac{dv}{dt} = ky [/itex] or

    [tex] v\frac{dv}{dy} = ky [/tex]

    My idea here was to get an expression for v, and then integrate that (as dy/dt) to get y in terms of t. But I have never really seen something like this before in the text. So perhaps I took a wrong turn here. What I tried to do was this: [tex] v\, dv = ky\, dy [/tex] and integrate this to get [tex] v^2/2 = ky^2/2 + C [/tex].

    I know the book says you aren't supposed to view dv/dy (or any derivative) as a fraction, but I couldn't think of anything else to do.

    On the off chance this was ok. I found [itex] C = v_0^2/2 [/itex] and so end up with [itex] v^2 = ky^2 + v_0^2 [/itex].

    So (assuming this is right), I have [tex] \frac{dy}{dt} = \sqrt{ky^2 + v_0^2} [/tex] I use the theorem on reciprocals to write it as [tex] \frac{dt}{dy} = \frac{1}{\sqrt{ky^2 + v_0^2}} [/tex]

    To integrate this, I set [itex] y = v_0\tan\theta/\sqrt{k} [/itex] and so [itex] dy/d\theta = v_o\sec^2\theta/\sqrt{k} [/itex] and I end up with the integral

    [tex] t = \frac{1}{\sqrt{k}}\int \frac{1}{\cos\theta}\, d\theta = \frac{\log(\cos\theta)}{\sqrt{k}} = \frac{1}{\sqrt{k}}\log\left(\frac{v_0}{\sqrt{v_0^2 + y^2k}}\right) + C [/tex]

    So

    [tex] De^{\sqrt{k}t} = \frac{v_0}{\sqrt{v_0^2+y^2k}} [/tex]

    and

    [itex] D = 1/v_0 [/itex].

    Here is where I stopped, since it is obvious that this is not going to lead to the correct answer as given in the book, which is

    [tex] y = \frac{v_0}{2\sqrt{k}}(e^{\sqrt{k}t} - e^{-\sqrt{k}t}) [/tex]

    Thanks again, if anyone has the fortitude to slog through all this.

    Sheldon
     
    Last edited: Feb 18, 2007
  2. jcsd
  3. Feb 18, 2007 #2
    Now I see a mistake in my integration of 1/cos. Let me think on this some more, and I will post again. Sorry to bother everyone.

    Sheldon
     
    Last edited: Feb 18, 2007
  4. Feb 19, 2007 #3
    Yes, that was the problem. Well, I managed on my own for once. Sorry to bother anyone who read this.

    Thanks
    Sheldon
     
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