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Acceleration in terms of displacement

  1. Apr 8, 2010 #1
    1. The problem statement, all variables and given/known data

    a)find the velocity when x = 3, b)find the time taken to travel between x=1 to x=3.

    when x = 1, v = 0

    2. Relevant equations


    3. The attempt at a solution

    a)using the chain rule I got v = 2x^2 + u (u as a constant), but I think that's wrong.
    i tried using dx/dt = dv/dt * dx/dv
    b)I integrated that with respect to time giving 3/16, but I think the solution there is wrong too.

    I'm not really sure how to tackle this type of problem, and I'm pretty sure I'm missing something here, my book hasn't shown me any examples so far relating to this so I hope someone can help me, I'm a bit rubbish at calculus anyway and there's no answer in the back to guide me.
  2. jcsd
  3. Apr 8, 2010 #2


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    Homework Helper

    It was a good start, but the result is not correct. The original differential equation is


    which can be transformed to


    but the integration results in


    Use the condition x=1, v= 0 to get u. (u=-2, so v^2=2(x^2-1). Then you can find v when x=3.

    For question b, you need the x(t) function. v=dx/dt.

    [tex]\frac{dx}{dt}=\sqrt{2(x^2-1)}\rightarrow \frac{1}{\sqrt{2}}\int_1^3{\frac{dx}{\sqrt{x^2-1}} = t[/tex].

  4. Apr 9, 2010 #3
    thanks that was a big help for me, so I really appreciate it, I worked through it again and I understand it now.
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