# Velocity as function of Displacement to Displacement as function of Time

1. Dec 28, 2008

### StephenSF8

I've measured velocities of a particle at varying displacements and characterized the velocity as $$V(x) = 0.0002x^2 - 0.6484x + 885$$.

You can see that I know velocity (V) as a function of displacement (x). Ultimately I want to end up with a function for displacement as a function of time (t). I imagine that somehow a chain rule is used to change the variables, but I'm having trouble figuring it out. The books I have glaze over the issue of non-constant acceleration...

Will somebody with more calculus experience help me out? Thanks.

Oh, and the initial conditions are x = 0, and so V(0) = 885.

2. Dec 28, 2008

### HallsofIvy

Since v= dx/dt you have, effectively, the differential equation $dx/dt= 0.0002x^2 - 0.6484x + 885$ which you can rewrite
[tex]\frac{dv}{0.0002x^2 - 0.6484x + 885}= dt[/itex]
Factor the denominator and use "partial fractions" to integrate.

3. Dec 28, 2008

### LennoxLewis

Shouldn't that "dv" in the last term be a "dx" ?

4. Dec 29, 2008

yes.