What is the Solution to the Retarding Force Problem?

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The discussion centers on the retarding force problem involving a particle moving in a medium influenced by the force mk(v^3 + va^2), where k and a are constants. It is established that the particle will never exceed the distance of (π)/(2ka) and will come to rest as time approaches infinity. The solution involves integrating the retarding force and applying trigonometric substitution, specifically v = acot(θ), to derive the time equation t = (-1/ka^2)(ln(v/(sqrt(v^2 + a^2))) + C. The participant confirms the correctness of their approach after verifying the differential equation.

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The Problem:
A particle moves in a medium influenced by a retarding force mk(v^3+va^2), where k and a are constants.
Show that
for any initial velocity the particle will never move more than (pi)/(2ka)
and
that it comes to rest only for t -> infinity

Attempt to solution:

I know I have to integrate the retarding force [F=-mk(v^3+va^2))] so that I can solve for v. Later, I have to solve for the constant C of the indefinite integral. Afterwards, I have to integrate to find the position.

To show the first part, I believe I have to find the value of the position as t-> infinity.
To show the second part, I believe I have to find the value of the velocity as t->infinity.

My problem: I cannot solve for v

I have solved the integral of the retarding force using trig substitution (v=acot(theta)) (dv=-acsc^2(theta)d(theta))
My final answer is:

t=(-1/ka^2)(ln(v/(sqrt(v^2+a^2))+C

If I use the quadratic equation to solve for v, I get imaginary numbers...
Am I making a mistake in my calculus or algebra?
or
Am I missing a physical concept?
 
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Draw a free body diagram for the particle in motion.
Apply ##F=m\dot v##.
 
After the free body diagram this is what I obtained.
Since the only force acting on it is the retarding force, it equals: F= -mk(v^3+va^2) = m(dv/dt)
This is a velocity dependent force. Therefore, I integrated the force as dt = -dv/(k(v^3+va^2)
then, I used trig substitution for v = acot(x) and dv=-csc^2(x)
And after integrating I obtain:

t= (-1/ka2)(ln(v/(sqrt(v^2 +a^2)) + C

Is this correct so far?
 
skeer said:
After the free body diagram this is what I obtained.
Since the only force acting on it is the retarding force, it equals: F= -mk(v^3+va^2) = m(dv/dt)
This is a velocity dependent force. Therefore, I integrated the force as dt = -dv/(k(v^3+va^2)
then, I used trig substitution for v = acot(x) and dv=-csc^2(x)
And after integrating I obtain:

t= (-1/ka2)(ln(v/(sqrt(v^2 +a^2)) + C

Is this correct so far?
Have you checked that by differentiating to get the ODE?
 
haruspex said:
Have you checked that by differentiating to get the ODE?
Yes, I checked. I realize also what was my problem. Thank you all
 

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