Finding displacement from variable acceleration

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SUMMARY

The discussion focuses on deriving the relationship between displacement (x) and velocity (v) under variable acceleration defined by the equation a = m - nv². The user is tasked with showing that after traveling a distance x, the car will achieve a velocity v. The relevant equations include vt = sqrt(m/n) for terminal velocity and x = -0.5ln(1 - (v/vt)²) for displacement. The solution involves integrating the equation vdv/(m - nv²) = dx to find the displacement in terms of velocity.

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  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concepts of variable acceleration and terminal velocity.
  • Knowledge of differential equations and their applications in physics.
  • Basic understanding of logarithmic functions and their properties.
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  • Study the integration of differential equations in physics contexts.
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  • Explore the concept of terminal velocity and its implications in fluid dynamics.
  • Investigate the application of logarithmic functions in solving physical problems.
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TPHughes
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Hello, I've been staring at this question for an hour with no luck.. if anyone could point me in the right direction that would greatly appreciated!

Homework Statement



a = acceleration
v = velocity
x = displacement
m and n are constants
vt = terminal velocity

Show that after traveling x, car will have velocity v.


Homework Equations



a = m - nv2 <- given equation

and vt = sqrt(m/n)

x = -0.5ln(1 - (v/vt)2) <- equation need to find

The Attempt at a Solution



I have attempted to substitute in a = d/dx(0.5v2) and then integrate with respect to x but had no luck.

Am stuck with where to go next! Any help would be greatly appreciated!
 
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Welcome to PF!

Hello TPHughes! Welcome to PF! :smile:
TPHughes said:
a = m - nv2 <- given equation

vdv/(m - nv2) = dx :wink:
 

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