Velocity as function of Displacement to Displacement as function of Time

[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.

 

[HallsofIvy]

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

 

[LennoxLewis]

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

 

[uq_civediv]

yes.