Position-time equation from force-position equation

  1. I've been trying to obtain an equation of position in terms of time given force in terms of position. I've tried and I think I've managed to obtain an equation of velocity in terms of position using work and kinetic energy but I haven't managed the position time equation.

    This is how I got velocity equation:
    A point mass [itex]m[/itex] is at starting position [itex]x=0[/itex] with starting velocity [itex]v_0[/itex]

    [itex]F=x+1[/itex]
    [itex]W=\int_0^x {(x+1)dx}=\frac{x^2+2x}2[/itex]
    [itex]\frac 1 2 m(v^2-{v_0}^2)=\frac{x^2+2x}2[/itex]
    [itex]v=\sqrt{\frac{x^2+2x}m+{v_0}^2}[/itex]

    How do I get a position-time equation? And how do I use a starting position other than x = 0?
     
  2. jcsd
  3. WannabeNewton

    WannabeNewton 5,862
    Science Advisor

    Well just note that your equations are horribly dimensionally incorrect. But other than that, if you have ##\dot{x} = f(x)## then ##t = \int _{0}^{x} \frac{1}{f(x)}dx## and from there you just have to integrate (in principle) and invert the equation (in principle) to get ##x(t)## explicitly.
     
  4. Sorry I forgot to mention that I was working in one dimension. And what does the dot above the x signify?
     
  5. WannabeNewton

    WannabeNewton 5,862
    Science Advisor

    When I said dimensionally incorrect I meant that your units are all wrong. The dot is a time derivative so ##v = \dot{x}##.
     
  6. By inverting the equation, do you mean getting an equation of x in terms of t?
     
  7. WannabeNewton

    WannabeNewton 5,862
    Science Advisor

    Yep.
     
  8. But does the velocity equation have to have a time variable in it? And also, is it possible to do this from an acceleration equation?
     
  9. WannabeNewton

    WannabeNewton 5,862
    Science Advisor

    No the velocity equation doesn't have to have a time variable. As I said, you have ##\frac{\mathrm{d} x}{\mathrm{d} t} = f(x)## hence you can solve for ##t(x)## just by solving the differential equation (in principle). You can then get ##x(t)## by inverting the equation (again in principle). I don't know what you mean by your second question; isn't that exactly what you did here? You started with the force as a function of position, which is also acceleration as a function of position, and you used the work-energy theorem to get an equation for the velocity in terms of position.
     
  10. I mean, is there was a way to do it without going through the process of obtaining a velocity equation?
     
  11. WannabeNewton

    WannabeNewton 5,862
    Science Advisor

    Well sure you could directly solve the second order linear differential equation in principle. The simplest example is the simple harmonic oscillator wherein you have ##\ddot{x} = f(x) = -\omega^{2} x##. You can solve this analytically immediately and obtain ##x(t) = A\cos\omega t + B\sin\omega t##.
     
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