Kinmatic question involving differential equation

AI Thread Summary
A particle of mass m falls from rest, experiencing air resistance proportional to the square of its speed, represented by kv². The problem requires proving that the distance fallen s can be expressed as s = V²/2g ln(V²/(V²-v²), where V² = mg/k. The discussion involves deriving the time equation t = V/2g ln((V+v)/(V-v)) and suggests integrating the velocity function v(t) to find s(t). Alternatively, the original differential equation can be transformed by treating s as the independent variable, leading to a separable equation that can be solved with initial conditions. The key focus is on finding the relationship between distance fallen and velocity through integration and transformation techniques.
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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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