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Suppose an objects starts with velocity $v$. A damping force that is dependent of velocity acts on this object ($F=-\beta v$). After traversing distance $l$, the object has velocity $\alpha v$ with $0<\alpha<1$. Calculate the drag coefficient $\beta$.

Well, it seems quite tricky. I tried starting with the dynamical equation: $$ ma=m\frac{dv}{dt}=-\beta v$$which we can rearrange to get $$ \frac{dv}{v}=-\frac{\beta}{m}dt$$We can integrate this: $$ \int_{v}^{\alpha v}\frac{dv}{v}=-\int_{0}^{t}\frac{\beta}{m}dt$$However, the answer is dependent of time. I also tried writing $$ ma=m\frac{dv}{dt}=m\frac{dv}{dx}\frac{dx}{dt}=m\frac{dv}{dx}v$$ and the dynamical equation becomes: $$ m\frac{dv}{dx}=-\beta$$ we can rearrange this: $$ mdv=-\beta dx$$ but again, after integrating, the answer is dependent in turn on the distance the object traversed.

Any hints?

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closed as off-topic by Qmechanic Jun 19 '17 at 19:59

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  • $\begingroup$ The velocity will get any value in $(0,v]$. This problem can't be solved. $\endgroup$ – AHB Jun 19 '17 at 19:39
  • $\begingroup$ OMG. I didn't see "After traversing distance l" ! And yes. It can be easily solved. Sorry for giving uneducated comment. $\endgroup$ – AHB Jun 19 '17 at 20:08
  • $\begingroup$ The question was edited by the OP to replace "after an unknown time" to "after traversing a distance $l$. The new problem can be solved, the old one cannot. $\endgroup$ – mike stone Jun 20 '17 at 13:53
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The solution to your equation is is $v(t)=v(0) \exp\{-\beta t/m\}$ as may be verified by plugging it in.

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  • $\begingroup$ Aaaand? It's dependent on time. The task is to express $\beta$ as a function of $v$ and $\alpha$ only. $\endgroup$ – Mermon Jun 19 '17 at 19:29
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    $\begingroup$ There is no solution to your original problem. You have not been given enough data. $\endgroup$ – mike stone Jun 19 '17 at 19:30
  • $\begingroup$ @Mermon In the part you quoted of the question it doesn't specify that it needs to be a function of only $v $ and $\alpha $. Are you sure you can't have $t $ in the answer? This question is about a transient response, without using time we can really solve it. $\endgroup$ – JMac Jun 19 '17 at 20:00
  • $\begingroup$ Well, and if we add that the body traversed distance $l$? $\endgroup$ – Mermon Jun 19 '17 at 20:02
  • $\begingroup$ NMevermind, solved it. $\endgroup$ – Mermon Jun 19 '17 at 20:06

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