## Finding an expression for x(t)

Suppose I have a particle on a line, and I know some function a(x) and the initial x, v, and a. How could I work out x(t)?

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 integrate! v(t) = integral of a(t)dt + v(t=0) x(t) = integral of v(t)dt + x(t=0)
 Thanks for the reply. From what I understand your solution requires me to know a(t), but what can I do if I only know a(x)?

## Finding an expression for x(t)

Ah, then you have:

x''(t) = f(x(t))

This is a 2nd order autonomous differential equation with a general solution, although if f is tricky then you might need a numerical integrator to solve it. The trick is to multiply through by 2x'(t) then factor the left side into ((x'(t))^2)'. Then you can find x(t) by integrating, taking the square root, and integrating again.

 Thanks a lot! I think I've seen that method applied before to simple harmonic motion, I wish I made the connection earlier.
 No problem. It caught me a little off guard as well!
 If you only know a(x), you can use the following: $$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v \frac{dv}{dx} = a(x)$$ Then: $$\int v dv = \frac{v^2}{2}= \int a(x) dx$$ After integrating, this can be solved for v(x). Then: $$\frac{dx}{dt} = v(x)$$ $$\int dt = t = \int \frac{dx}{v(x)}$$ After integrating, this gives t(x), which can then be inverted to give x(t). This procedure can be complicated and mathematically difficult, but it will work, at least numerically.
 Thank you, that's exactly what I was looking for
 [deleted] I just repeated the same thing that phyzguy said.

 Tags acceleration, calculus, particle, position