Velocity as function of Displacement to Displacement as function of Time

AI Thread Summary
The discussion centers on deriving a displacement function from a given velocity function, V(x) = 0.0002x^2 - 0.6484x + 885. The user seeks assistance in applying calculus, particularly the chain rule, to convert the velocity as a function of displacement into displacement as a function of time. They note the initial conditions of x = 0 and V(0) = 885, leading to the differential equation dx/dt = 0.0002x^2 - 0.6484x + 885. The user acknowledges a mistake in notation, confusing "dv" with "dx," indicating a need for clarification on integrating the equation using partial fractions. Overall, the thread highlights the challenges of dealing with non-constant acceleration in calculus.
StephenSF8
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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.
 
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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.
 
Shouldn't that "dv" in the last term be a "dx" ?
 
yes.
 

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