Velocity as function of Displacement to Displacement as function of Time

  • Thread starter StephenSF8
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  • #1
I've measured velocities of a particle at varying displacements and characterized the velocity as [tex]V(x) = 0.0002x^2 - 0.6484x + 885[/tex].

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.
 

Answers and Replies

  • #2
HallsofIvy
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Since v= dx/dt you have, effectively, the differential equation [itex]dx/dt= 0.0002x^2 - 0.6484x + 885[/itex] 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
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Shouldn't that "dv" in the last term be a "dx" ?
 
  • #4
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yes.
 

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