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Need help converting a function of displacement to a function of time.

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Assume we have some mass, m, that varies it's distance as v(x)=1/x^2. Assume v(x=0)=0 at t=0.

    What is the force F(x) that causes this motion?
    Find x(t).
    What is F(t)?

    2. Relevant equations
    [tex]F=m\ddot{x}[/tex]


    3. The attempt at a solution
    I think I need to take the derivative of my velocity with respect to x to find my acceleration. With that I can find the force as a function of x? I think this may be wrong, but I'm really not sure. Could someone help me out as to where to begin?
     
  2. jcsd
  3. Oct 2, 2008 #2

    Hootenanny

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    HINT: By the chain rule

    [tex]a = \frac{dv}{dt} = \frac{dv}{dx}\overbrace{\frac{dx}{dt}}^{v}[/tex]

    [tex] \Rightarrow a = v\frac{dv}{dx}[/tex]
     
  4. Oct 2, 2008 #3
    By this, I find the following, by the way... in the problem I'm solving v(x)=ax^-3, where a is a constant.

    [tex]v(x)=ax^{-3}[/tex]
    [tex]F(x)=m\ddot{x}[/tex]
    [tex]F(x)=m\frac{dv}{dt}[/tex]
    [tex]F(x)=m\frac{dv}{dx}\frac{dx}{dt}[/tex]
    [tex]\frac{dv}{dx}=-3ax^{-4}[/tex]
    [tex]F(x)=m*-3ax^{-4}*ax^{-3}=-3ma^2x^{-7}[/tex]
    [tex]F(t)=m\frac{dv}{dt}=-3ma^2x^{-7}[/tex]
    [tex]\frac{dv}{dt}=-3a^2x^{-7}[/tex]
    [tex]dv=-3a^2x^{-7}dt[/tex]
    [tex]\int dv=\int -3a^2x^{-7}dt[/tex]
    [tex]v(t)=\frac{dx}{dt}=-3a^2x^{-7}t[/tex]
    [tex]x^7 dx=-3a^2t dt[/tex]
    [tex]\int x^7 dx=\int -3a^2t dt[/tex]
    [tex]\frac{x^8}{8}=\frac{-3}{2}a^2t^t[/tex]
    [tex]x^8=-12a^2t^2[/tex]
    [tex]x(t)=(12a^2t^2)^{1/8}[/tex]
     
  5. Oct 2, 2008 #4

    Hootenanny

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    You're good upto here, after this things get a bit hairy. When integrating don't forget that x is a function of time, therefore you should note that

    [tex]\int x(t) dt \neq xt+ const.[/tex]

    Notice that the question asks you to determine x(t) before finding F(t), therefore it might be easier to find x(t) next.
     
  6. Oct 2, 2008 #5
    So to find x(t)?:

    [tex]v(x)=ax^{-3}=\frac{dx}{dt}[/tex]
    [tex]x^3 dx = a dt[/tex]
    [tex]\int x^3 dx = \int a dt[/tex]
    [tex]\frac{x^4}{4}=at[/tex]
    [tex]x(t)=(4at)^{1/4}[/tex]
     
  7. Oct 2, 2008 #6

    Hootenanny

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    Looks good to me :approve:
     
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