Velocity as function of Displacement to Displacement as function of Time

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SUMMARY

The discussion focuses on converting the velocity function V(x) = 0.0002x^2 - 0.6484x + 885, which describes the velocity of a particle as a function of displacement, into a displacement function as a function of time, x(t). The user seeks assistance with applying calculus techniques, particularly the chain rule and integration methods, to address the challenge of non-constant acceleration. Initial conditions are provided, with x = 0 and V(0) = 885, leading to the differential equation dx/dt = 0.0002x^2 - 0.6484x + 885. The discussion emphasizes the need for proper integration techniques, including the use of partial fractions.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and integration.
  • Familiarity with differential equations and their applications.
  • Knowledge of the chain rule in calculus.
  • Experience with partial fraction decomposition in integration.
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  • Study the application of the chain rule in transforming variables in calculus.
  • Learn techniques for solving differential equations, particularly non-constant acceleration scenarios.
  • Explore integration methods, focusing on partial fractions for complex rational functions.
  • Investigate the relationship between velocity, displacement, and time in physics, particularly in kinematics.
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Students and professionals in physics and engineering, particularly those dealing with kinematics and differential equations, as well as anyone seeking to deepen their understanding of calculus applications in motion analysis.

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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