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Position-time equation from force-position equation

by Droctagonopus
Tags: equation, forceposition, positiontime
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Droctagonopus
#1
Aug6-13, 03:44 AM
P: 30
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?
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WannabeNewton
#2
Aug6-13, 04:16 AM
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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.
Droctagonopus
#3
Aug6-13, 04:20 AM
P: 30
Sorry I forgot to mention that I was working in one dimension. And what does the dot above the x signify?

WannabeNewton
#4
Aug6-13, 04:21 AM
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Position-time equation from force-position equation

When I said dimensionally incorrect I meant that your units are all wrong. The dot is a time derivative so ##v = \dot{x}##.
Droctagonopus
#5
Aug6-13, 04:28 AM
P: 30
By inverting the equation, do you mean getting an equation of x in terms of t?
WannabeNewton
#6
Aug6-13, 04:28 AM
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Yep.
Droctagonopus
#7
Aug6-13, 04:31 AM
P: 30
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?
WannabeNewton
#8
Aug6-13, 04:33 AM
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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.
Droctagonopus
#9
Aug6-13, 04:36 AM
P: 30
I mean, is there was a way to do it without going through the process of obtaining a velocity equation?
WannabeNewton
#10
Aug6-13, 04:38 AM
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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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