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Variable acceleration problem

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    knowledge required: a= dv/dt=v dv/ds

    Particle moves along straight line with a = 5-2v v=0, t=0
    show that  t = \frac {1}{2} ln (\frac{5}{5-2v})s
    and hence express v as a function of t
    show that the velocity of the particle approaches a max value and find this max value
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    Could you post your attempts thus far?

    Edit: Are you sure the question information is correct? (Think the 's' in the solution shouldn't be there)
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    aah, only now I realise that that s will be for seconds! Let me try again and post my solution
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    Is there a way of saving the latex tags? Instead of having to type them out or copy and paste them?
    Anyway,
    \frac{dv}{dt} =5-2v

    t+c'=\int\frac{1}{5-2v}dv

    t+c'=\frac{-1}{2}\int\frac{1}{u}du

    t+c'=\frac{-1}{2}ln(5-2v)

    c=\frac{-1}{2}ln(5-2(0))=\frac{-1}{2}ln5

    t=\frac{-1}{2}ln(5-2v)+\frac{1}{2}ln5
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    That's the same as what they want, but obviously not simplified. It's just laws of logs from there though.
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    I don't see it,
    I end up with t=\frac{1}{2}ln\frac{5}{5-2v}-\frac{1}{2}
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t = -\dfrac{1}{2} ln(5-2v) + \dfrac{1}{2}ln5 \\





= \dfrac{1}{2}( ln\dfrac{1}{5-2v} + ln5)\\





=\dfrac{1}{2} ln \dfrac{5}{5-2v}

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Updated: July 18, 2012
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