# Motion with friction depending on velocity and distance travelled. (not oscillation)

I am a mathematician by trade and I've stumbled upon some physics problems that I should be able to do, but just cannot. Anybody else in my predicament? Anyway.

Consider this vaguely stated problem. A mass starts moving linearly with initial velocity $$v_0$$, and is subjected to a friction force of $$F=bxv$$, $$b>0$$ constant, $$x=x(t)$$ is the distance traveled and $$v(t)$$ is the velocity. Find the distance it will travel and the total time of motion.

So, if I am getting this right, $$v(t)=x'(t)$$. I cannot try conservation of energy but Newton's second law of motion is applicable, right? So I get the ODE $$bx(t)x'(t)=mx''(t)$$ Solving this gives

$$bx^2(t)/2=mx'(t)+c \ \ \ \ \ (1)$$ Now there's my issue #1. If I let $$x(0)=0$$, then $$c=-mv_0$$. Then I get the differential equation $$mx'(t)=bx^2(t)+mv_0$$ which I can integrate to get

$$x(t)=\frac{b^{1/2}}{(mv_0)^{1/2}}tan\left(\frac{(mv_0)^{3/2}}{b^{1/2}}t\right)$$

This seems wrong along so many lines ha,ha

How do I retrieve the total distance traveled this way? It seems like this motion will not stop.

Should I have chosen $$x(0)=x_0$$ so that (1) gives me a negative constant c? Then I would have a totally different $$x(t)$$, I think. Then how would I get the distance traveled, since it would also depend on t?

• A friction force is against the direction of v. You have them going in the same direction. Fix that and things will make more sense. – Bill Watts Nov 2 '19 at 0:24
• Your force 1) makes the particle go faster, 2) increases with increasing velocity, and 3) increases as the particle gets further from the origin. Given this, the fact that its position goes to infinity in finite time isn't entirely surprising. – probably_someone Nov 2 '19 at 1:50

For a frictional force (i.e. one that slows down the particle), $$b$$ must be negative. When you set $$b<0$$, you obtain:
$$x(t)=\frac{i|b|^{1/2}}{(mv_0)^{1/2}}\tan\left(\frac{(mv_0)^{3/2}}{i|b|^{1/2}}t\right)=\frac{|b|^{1/2}}{(mv_0)^{1/2}}\tanh\left(\frac{(mv_0)^{3/2}}{|b|^{1/2}}t\right)$$
since $$\tan\left(\frac{k}{i}t\right)=\frac{1}{i}\tanh(kt)$$. This means the particle never actually stops moving (but that was never guaranteed anyway), but effectively comes to rest after a long time at the position
$$x_{stop}=\lim_{t\to\infty}x(t)=\frac{|b|^{1/2}}{(mv_0)^{1/2}}\frac{\pi}{2}$$