Kinmatic question involving differential equation

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SUMMARY

The discussion focuses on solving a differential equation related to the motion of a particle falling under the influence of gravity and air resistance. The equation derived is s = V²/2g ln(V²/(V²-v²)), where V² = mg/k, representing the terminal velocity. The user has successfully proven the time equation t = V/2g ln((V+v)/(V-v)) and seeks guidance on how to utilize this to derive the distance fallen, s. Two methods are proposed: integrating the velocity function v(t) to find s(t) or transforming the differential equation to express it in terms of the distance fallen, s, using the chain rule.

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  • Understanding of differential equations and their applications in physics.
  • Familiarity with the concepts of air resistance and terminal velocity.
  • Knowledge of calculus, specifically integration and the chain rule.
  • Basic principles of mechanics, including Newton's laws of motion.
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  • Study the derivation of the equation for terminal velocity in fluid dynamics.
  • Learn about the method of separation of variables in differential equations.
  • Explore the application of the chain rule in physics problems involving motion.
  • Investigate numerical methods for solving complex differential equations.
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Students studying physics, particularly those focusing on mechanics and differential equations, as well as educators seeking to enhance their understanding of motion under resistance.

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Homework Statement


A particle of mass m falls fom rest; the resistance of the air when the speed is v is kv2 where k is a constant. If s is the distance fallen in time t, prove that s = V2/2g ln(V2/(V2-v2)), where V2=mg/k


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The Attempt at a Solution


I have already proven that t = V/2g ln((V+v)/(V-v)). How do I make use of that equation to solve the problem? Do I have to change to v(t) equation and integrate it?
 
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Yes, this is one way. Integrating v(t) you get s(t), but you need the s(v) relation, so you have to substitute t(v) for t.

The other way is that you transform the original differential equation:

dv/dt= g-k/m * v^2

by considering s the independent variable and applying the chain rule during differentiation with respect to t.

dv/dt=dv/ds*ds/dt = dv/ds *v .

The new differential equation is : v dv/ds = g-k/m*v^2.
This is separable, easy to solve with the condition that v=0 at s=0.



ehild
 

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