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Solve differential a=kxv for x(t) and x(v)

  1. Aug 15, 2014 #1
    1. The problem statement, all variables and given/known data
    If the acceleration of a particle is given by a=k*x*v , where x is position, v is velocity and k is constant and positive, and in t=0: x(t=0)=0 and v(t=0)=vo; find x(t) and x(v).


    3. The attempt at a solution

    I arrived to dx/dt= k/2*(x2-xo)+vo , (xo=x(t=0)=0). Then I tried to solve the (definite) integral (from 0 to x) ∫1/((k/2)*(x2)+vo)dx that is equal to ∫dt = t. In this integral (which I just couldn't solve) I took vo as a constant. My idea was to solve (result of the integral)=t for x(t).
    Is this the right way to do it? If so, how can I solve the integral?

    Thanks!
     
  2. jcsd
  3. Aug 15, 2014 #2

    pasmith

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    Homework Helper

    Having obtained [tex]
    \frac{d\dot x}{dx} = kx
    [/tex] it's easier to set [tex]
    \dot x = \tfrac12 k (x^2 + C)
    [/tex] where [itex]C = 2v_0/k[/itex] so that [tex]
    \int \frac{1}{x^2 + C}\,dx = \tfrac12 kt.
    [/tex]

    To do the integral, you will want to make a suitable substitution so that you can use
    [tex]
    \int \frac{1}{1 + u^2}\,du = \arctan(u) + D
    [/tex]
     
  4. Aug 15, 2014 #3
    Then for x(v) can we take equation v=k/2*x2+vo and so for x : x(v)=2(v-vo)/k ???
     
  5. Aug 15, 2014 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, x is squared. You should have [tex]x= \sqrt{2(v- v_0)/k}[/tex]
     
  6. Aug 15, 2014 #5
    Oh, of course, silly mistake.. Thanks!
     
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