This is a mechanics question but is pretty much mathematical so I figured I should post it here.

If I had a particle dropped from rest and it had resistance $mkv$ where mass is $m$, $v$ is velocity, and $k$ is a constant, how would the velocity-time graph change if the resistance was given by $mkv^2$ instead?

I know both would approach terminal velocity but how would the shape change?


First of all by newtons second law we have choosing the coordinate axis as down positive and up as negative we have the following equation for the first one.

$F = mg - mkv = ma \rightarrow g - kv = a \rightarrow g - kv = dv/dt$ Integrate and we will get the graph second case we do something similar.

Now for the second equation we have $F = mg - mkv^2 = ma$


First case: From Newton's Second Law, you have: $$F=ma=mg-mkv\Longrightarrow\dfrac{dv}{dt}=g-kv$$ We can perform an integration using separation of variables: $$\int_0^v\dfrac{dv}{g-kv}=\int_0^t dt\Longleftrightarrow -\dfrac{1}{k}\ln{|g-kv|}\Bigg\vert_0^v=t\Bigg\vert_0^t$$ Rearranging: $$v(t)=\dfrac{g}{k}\left(1-e^{-tk}\right)$$ And you can see that: $$v_{\text{terminal}}=\lim_{t\to\infty}v(t)=\dfrac{g}{k}$$

Second case: From Newton's Second Law, you have: $$F=ma=mg-mkv^2\Longrightarrow \dfrac{dv}{dt}=g-kv^2$$ Using separation of variables again: $$\int_0^v\dfrac{dv}{g\left(1-\dfrac{kv^2}{g}\right)}=\int_0^t dt$$ Making the substitution $u=\dfrac{kv^2}{g}$, the integral becomes: $$\dfrac{1}{\sqrt{gk}}\int_0^v\dfrac{du}{1-u^2}=\int_0^t dt\Longleftrightarrow\dfrac{\tanh^{-1}\left(v\sqrt{\frac{k}{g}}\right)}{\sqrt{gk}}\Bigg\vert_0^v=t\Bigg\vert_0^t$$ Rearranging: $$v(t)=\sqrt{\dfrac{g}{k}}\tanh\left(\sqrt{gk}t\right)$$ And you can see that: $$v_{\text{terminal}}=\lim_{t\to\infty}v(t)=\sqrt{\dfrac{g}{k}}$$


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