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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?

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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.

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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! –  Steven Harris Oct 14 '12 at 19:00
    
No problem. It's not just $\alpha<0$ though! For example, try $\alpha=\frac{1}{2}$. –  AClassicalCaseOfConfusion Oct 15 '12 at 7:42
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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.

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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. –  Prathyush Oct 14 '12 at 9:12
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