# Velocity for t->infinity with given equation of motion

1. Dec 12, 2015

### spacetimedude

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. 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!

2. Dec 12, 2015

### nrqed

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.

3. Dec 12, 2015

### PeroK

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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