Question

If the differential equation of a body of mass $m$ falling from rest subjected to the force of gravity and an air resistance proportional to the square of the velocity is given by $mv\frac{dv}{dx}=k{a}^2-k{v}^2$, then it can be proved that $\frac{2kx}{m}=\log \left(\frac{a^2}{a^2-v^2}\right)$, where $mg=k{a}^2$

• A. True
• B. False

Answer 1

The correct option is A True

We have $mv\frac{dv}{dx}=k{a}^2-k{v}^2=k(a^2-v^2)$

This is a differential equation of variable separable type.

$\therefore \frac{v}{a^2-v^2}dv=\frac{k}{m}dx$ ,,,(1)

By integration, $-\frac{1}{2}\log (a^2-v^2)=\frac{k}{m}+\log c$

But by data when $t=0,x=0,v=0$.

Therefore, $\log c=-\frac{1}{2}\log a^2$

$\Rightarrow -\frac{1}{2}\log (a^2-v^2)=\frac{kx}{m}-\frac{1}{2}\log a^2$

$\Rightarrow \frac{2kx}{m}=\log a^2-\log (a^2-v^2)$

$\Rightarrow \frac{2kx}{m}=\log \left(\frac{a^2}{a^2-v^2}\right)$