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F(v) related to x(t)

  1. Oct 11, 2006 #1
    A particle of mass m is subject to a force F(v) = -m(alpha)v^2. The initial position is zero, and the initial speed is v nought. Find x(t).
     
  2. jcsd
  3. Oct 11, 2006 #2

    quasar987

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    It's only a matter of solving the differential equation

    [tex]-m\alpha v^2=m\frac{dv}{dt}[/tex]

    and then

    [tex]v=\frac{dx}{dt}[/tex]
     
  4. Oct 11, 2006 #3
    Sorry perhaps I'm not catching on, so I have:

    -1/v = -(alpha)t

    Do I merely say that 1/(alpha t) = dx/dt or perhaps I'm missing something.
    Or is my first step incorrect?
     
  5. Oct 11, 2006 #4

    quasar987

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    You forgot the constant of integration. The solution to the differential equation for v is

    -(alpha)t = -1/v +C

    And plugging v(0)=[itex]v_0[/itex] gives C=1/[itex]v_0[/itex]. So

    [tex]v(t)=\frac{1}{\alpha t}+\frac{1}{v_0}[/tex]

    And now your have to solve the differential equation.

    [tex]\frac{dx}{dt}= \frac{1}{\alpha t}+\frac{1}{v_0}[/tex]

    with initial condition x(0)=0 to find x(t).

    Makes sense?
     
  6. Oct 11, 2006 #5
    Okay, so I got:

    x(t) = (ln(t)/alpha) + (t/v nought) + C
    and C = 0 at x(0)

    So I have x(t) = (ln(t)/alpha) + (t/v nought)

    Is that correct or did I once again miss something?
     
  7. Oct 11, 2006 #6

    quasar987

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    It looks fine.
     
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