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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 :)

Qmechanic
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Yotam Salmon
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1 Answers1

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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.

Avantgarde
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Rishabh Jain
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