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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
  2. Offline

    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)
  3. Offline

    aah, only now I realise that that s will be for seconds! Let me try again and post my solution
  4. Offline

    Is there a way of saving the latex tags? Instead of having to type them out or copy and paste them?
    \frac{dv}{dt} =5-2v





  5. Offline

    That's the same as what they want, but obviously not simplified. It's just laws of logs from there though.
  6. Offline

    I don't see it,
    I end up with t=\frac{1}{2}ln\frac{5}{5-2v}-\frac{1}{2}
  7. Offline


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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