Velocity for t->infinity with given equation of motion

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SUMMARY

The discussion focuses on the motion of a unit mass body subjected to a velocity-dependent force described by the equation F = a(dx/dt) - b(dx/dt)^2. The equation of motion is derived as x'' - ax' + b(x')^2 = 0. As time approaches infinity, the velocity asymptotically approaches the fixed point v = a/b, where acceleration becomes zero, indicating that the body reaches a terminal velocity. This conclusion is supported by analyzing the acceleration as a function of velocity, revealing that acceleration is positive for velocities less than a/b, zero at a/b, and negative for velocities greater than a/b.

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Homework Statement


A body of unit mass, whose position is x(t) is subject to a velocity-dependent force of the form
F=a(dx/dt)-b(dx/dt)^2
where a and b are positive constants and the positive x direction is to the right.
a) Write down the equation of motion
b) If the motion is initially to the right, what would be the velocity for t->infinity?

Homework Equations

The Attempt at a Solution


Part a is straight forward. Since m is a unit mass, we can set m=1. Then we have F=(1)x''=ax'-b(x')^2 =>x''-ax'+b(x')^2=0.
But I do not know how to start with part b.
The solution says "the equation has fixed point at v=a/b so it will asymptotically reach a/b as t->infinity" but I am not sure how this came about. Why set x''=0 to find the velocity at t->infinity?
Thanks!
 
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spacetimedude said:

Homework Statement


A body of unit mass, whose position is x(t) is subject to a velocity-dependent force of the form
F=a(dx/dt)-b(dx/dt)^2
where a and b are positive constants and the positive x direction is to the right.
a) Write down the equation of motion
b) If the motion is initially to the right, what would be the velocity for t->infinity?

Homework Equations

The Attempt at a Solution


Part a is straight forward. Since m is a unit mass, we can set m=1. Then we have F=(1)x''=ax'-b(x')^2 =>x''-ax'+b(x')^2=0.
But I do not know how to start with part b.
The solution says "the equation has fixed point at v=a/b so it will asymptotically reach a/b as t->infinity" but I am not sure how this came about. Why set x''=0 to find the velocity at t->infinity?
Thanks!
As t goes to infinity, the velocity approaches an asymptotic value which, by definition, is a constant. In other words, the velocity approaches a certain value and does not change in time anymore, so its derivative (the acceleration) is then zero. This is the trick used to find, for example, the terminal velocity of a falling object: upon reaching the terminal speed, the acceleration is zero.
 
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spacetimedude said:
But I do not know how to start with part b.
The solution says "the equation has fixed point at v=a/b so it will asymptotically reach a/b as t->infinity" but I am not sure how this came about. Why set x''=0 to find the velocity at t->infinity?
Thanks!

It's probably a good idea to check for yourself why the asymptotic argument works in this case. If you draw a graph of acceleration against velocity (##acc = av - bv^2##) then you'll notice that for ##0 < v < a/b## acceleration is positive, for ##v = a/b## acceleration is 0 and for ##v > a/b## acceleration is negative.

If ##v_0 > 0##, then there are two cases: ##0 < v_0 < a/b## and ##v_0 > a/b##. You should try to figure out what happens to ##v## in those two cases, as ##t## increases.
 
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