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Advanced mechanics - x(t) from v(x)

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m's velocity varies according to bx-n

    Find the position as a function of time, setting x = x0 at t=0

    2. Relevant equations

    v(x) = bx-n

    possibly relevant: f(x) = -b2mnx-2n-1

    3. The attempt at a solution

    The first part of the question asked me to find the force acting on the particle as a function of x, which I did using the chain rule. I'm a little unclear as to whether I need f(x) to get x(t).

    Anyway, here's my attempt at a solution:
    dv/dt = (dv/dx)*(dx/dt)
    dv/dt = (dv/dx) * v(x)

    Both of these quantities are known so I plugged them in and got an expression for dv/dt. I then tried to integrate that expression twice, once to get v(t) and another time to get x(t). However, when I do that I just get the expression times t2/2, which would make x(0) = 0, not x0 as the problem statement gives.

    Am I doing this the complete wrong way or am I on the right track and just not understanding calculus?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 25, 2011 #2
    [tex]
    v = \frac{d x}{d t} = b x^{-n}
    [/tex]
    This is a 1st order ODE with separable variables. The variables separate as:
    [tex]
    x^{n} \, dx = b \, dt
    [/tex]
    which can be integrated by elementary tabular integrals. The one arbitrary constant is found from the initial condition.
     
  4. Sep 25, 2011 #3
    Thanks!

    I solved the ODE by integrating both sides and got xn+1/(n+1) = bt. I don't see where the constant comes into play here.
     
  5. Sep 25, 2011 #4

    rude man

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    Finish the expression for the integration of your ode. What must one add to every indefinite integral of df/dx?
     
  6. Sep 25, 2011 #5
    so I have:

    xn+1/(n+1) = bt + C

    then I plug in the initial condition x=x0 at t = 0

    C = x0n+1/(n+1)

    I just need to solve this for x now, correct?
     
  7. Sep 25, 2011 #6
  8. Sep 25, 2011 #7

    rude man

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    Molto bene, molto bene!
     
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