Question

105. Calc Air drag is a significant problem in some situations. Suppose the acceleration of a falling object is given by the following equation a(v) = g ? betav^2 (down is positive) where beta is a positive constant. (a) By integrating, find the velocity of a falling object as a function of time. (b) Find the terminal velocity of an object that falls from rest starting at t = 0.

Answer 1

$a(v)=g-\beta v^{2}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t}=g-\beta v^{2}=\beta (\frac{g}{\beta }-v^{2})$

$\Rightarrow \frac{dv}{(\frac{g}{\beta }-v^{2})}=\beta dt$

Let (g/B)=k².

Now , $\frac{1}{(k^{2}-v^{2})}=\frac{1}{2k}(\frac{1}{k+v}+\frac{1}{k-v})$

So,the differential equation becomes

$\frac{1}{2k}(\frac{1}{k+v}+\frac{1}{k-v})dv=\beta dt$

Integrate this equation with limits :

• v = 0 to v = v
• t = 0 to t = t

to get   $\frac{1}{2k}[\ln (k+v)-\ln (k-v)]=\beta t$

$\Rightarrow ln(\frac{k+v}{k-v})=2k\beta t$

$\Rightarrow (\frac{k+v}{k-v})=e^{2k\beta t}$

$\Rightarrow v=k [\frac{e^{2k\beta t}-1}{e^{2k\beta t}+1}]$ $\Rightarrow v=k [\frac{e^{2k\beta t}-1}{e^{2k\beta t}+1}]= k \tanh (k\beta t)$

Substitute k=sqrt(g/B) to get

$\Rightarrow v= \sqrt{\frac{g}{\beta }} \tanh (\sqrt{(g\beta )} t)$

The terminal velocity is that at which acceleration 'a' is zero and and object moves uniformly with this velocity.

$\Rightarrow a_{terminal}=g-\beta v_{terminal}^{2}=0$

$\Rightarrow v_{terminal}=\sqrt{\frac{g}{\beta }}$