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I've faced this question when trying to solve a test problem I had. The test, of course, did not require those functions to be expressed, but I just had to know what's going on.

We're given a body falling in the air. Air friction is defined so $|\vec f|=kv^2$, when $k$ is a constant, valued $0.25 \frac {kg} {m}$ and the mass is $10kg$.

Now, developing some equations;

$$\vec {\Sigma F _y}=kv^2-m \vec g$$

Then I know that the body will reach a constant speed when $\vec v=\sqrt {\frac {m \vec g} {k}}$.

I wanted to know how I can plot the velocity by time function. I tried to express the acceleration: $$a_{(v)}=\frac {\vec {\Sigma F _y}} {m}=\frac {kv^2} {m} - g$$

So now I have an acceleration by velocity function. Is it possible to develop it to a form of velocity by time function? Or, simpler, to an acceleration by time function?

I've tried to integrate the function, but then I realized that as it's $a_{(v)}$ and not $a_{(t)}$ it won't be of any help (Or will it? Does it have a physical meaning)

I'm stuck in my own thoughts with that one, would like to have some hints :)

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  • $\begingroup$ That you call it $a(v)$ suddenly in the last step seems arbitrary. It doesn't matter if you call it $a(t)$ or $a(v)$ (or $a(m)$ or $a(g)$ or... etc.). Writing $a(v)$ just emphasizes that $a$ depends on $v$. Which it does. That doesn't change the fact that it depends on $t$ as well. You can integrate over any of the dependent parameters that vary. But to continue in your derivation, I believe you will need some expression for $v$ that includes $t$, which you don't have. It might be a difficult task. $\endgroup$ – Steeven Jun 16 '17 at 20:51
  • $\begingroup$ But if I integrate $a(v)$ (or just $a$ is it?) over $v$ and not $t$, will it still be equal to the $v(t)$ or will it have a different meaning? I mean, is there a way to reach somehow the $t$ from all the mess here? $\endgroup$ – Yotam Salmon Jun 16 '17 at 20:55
  • $\begingroup$ 2D quadratic drag was also considered in this Phys.SE post and links therein. $\endgroup$ – Qmechanic Jun 16 '17 at 21:18
  • $\begingroup$ The point is that your expression is not wrong and the name $a(v)$ not important. You can integrate to $t$ if you want. But of course in order to carry that integration to $t$ all the way through, you must know how all the parameters in the expression depend on $t$. So, since $v$ depends on $t$ you have to insert that relation first. If you don't have it, you can't integrate to $t$ all the way. Then you will have to use some other method, if there is one. And I can see that the answer below actually shows how you can use a trick to avoid having to integrate to $t$ at all. $\endgroup$ – Steeven Jun 16 '17 at 23:44
  • $\begingroup$ Yeah, I understood you. But in order to integrate on $t$ to get the $v(t)$ i need to know first $v(t)$. Seems problematic a bit? ;) I'll try going on with the answer, but thanks anyways! $\endgroup$ – Yotam Salmon Jun 17 '17 at 10:04
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Acceleration is acceleration. It could be a function of time or velocity, doesn't matter. HINT- $a=\frac{d^2x}{dt^2}=\frac{d}{dx}\Big(\frac{dx}{dt}\Big)\times \frac{dx}{dt}=v\times \frac{dv}{dx}$ by chain rule. Now substitute this expression of $a$ in terms of $v$ and integrate which shouldn't be too difficult.

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  • $\begingroup$ Wait, wait. I lost you on the second $=$ (Sorry for my ignorance). Where is that theorem taken from? $\endgroup$ – Yotam Salmon Jun 16 '17 at 21:26
  • $\begingroup$ You mean the chain rule? it is very famous in differential calculus .You can just google it.For this particular derivation , go to quora.com/In-calculus-how-do-I-prove-that-a-v-*-dv-dx $\endgroup$ – Rishabh Jain Jun 16 '17 at 22:24

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