If a particle moves in a place with air resistance (but no other forces), will it ever reach a zero velocity in finite time? The air resistance is proportional to some power of velocity - $v^\alpha$, and I have to try it with different $\alpha$. I've solved for the function of position for several alphas and all functions I've gotten decay to $v=0$ as $t \to\infty$, but none ever reach exactly $v=0$ for a finite value of $t$. This should be the case, right?

  • $\begingroup$ "Air resistance", but "no other forces"?? Sooo... no brownian motion? Zero velocity.. please define what you mean by "zero velocity"? Is one planck distance per second "zero"?. there are a lot of assumptions needed before this can be answered satisfactorily. $\endgroup$
    – PcMan
    Dec 20, 2020 at 12:01

2 Answers 2


Yes. Without any force it indeed would reach zero speed only in $t=\infty$.

There is no contradiction with the real world. Here we have not only the resistance force, but more and more forces, and a dependance on size and shape of body.

  • 1
    $\begingroup$ Instead of zero and infinity, if you set "any" minimum velocity that you approximate as zero for practical considerations, then it would reach that velocity in a finite time. The problem with zero is that beyond a certain point Brownian motion will be significant. $\endgroup$
    – Prathyush
    Oct 14, 2012 at 9:12

Have you considered the full range of values of $\alpha$?

For $\alpha\ge1$, your conclusion is correct: the velocity approaches $v=0$ asymptotically at large times. If you consider $\alpha<1$, you can find solutions which reach $v=0$ in finite time. I'll leave the explicit solutions to you, but I find the time at which the particle stops to be

$$ T=\frac{v(0)^{1-\alpha}}{1-\alpha} \quad {\rm for}\quad \alpha<1. $$

Edit: I should just point out that my constant of proportionality was fixed to $1$. That is, I solved $v^\prime = -v^\alpha$. To get the answer in physical units, you'll have to reinstate it.

  • $\begingroup$ Ah, ok, I didn't even think of $\alpha$ < 0, because the 3 alphas that I had to try were 1 , 3/2 , 2. Thanks, though! $\endgroup$ Oct 14, 2012 at 19:00
  • $\begingroup$ No problem. It's not just $\alpha<0$ though! For example, try $\alpha=\frac{1}{2}$. $\endgroup$ Oct 15, 2012 at 7:42

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