Question

The equation of motion of a body falling under gravity is given by $\frac{dv}{dt}=g-\frac{g}{{\lambda }^2}{v}^2$. Distance travelled as a function of time is given by?. Give at $v=0$ ,$t=0$.

• A. $x=\frac{{\lambda }^2}{g}logsinh\left(\frac{gt}{\lambda }\right)$
• B. $x=\frac{\lambda }{g}logcosh\left(\frac{gt}{{\lambda }^2}\right)$
• C. $x=\frac{{\lambda }^2}{g}logsech\left(\frac{gt}{\lambda }\right)$
• D. $x=\frac{{\lambda }^2}{g}logcosh\left(\frac{gt}{\lambda }\right)$

Answer 1

Answer: (D)

$x=\frac{{\lambda }^2}{g}logcosh\left(\frac{gt}{\lambda }\right)$

We have $\frac{dv}{dt}=g-k{v}^2$

Now taking $k=\frac{g}{{\lambda }^2}$ for convenience

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

This is a differential equation of variable separable type.

$\therefore \frac{dv}{{\lambda }^2-v^2}=\frac{g}{{\lambda }^2}=\frac{g}{{\lambda }^2}dt$

By integration, $\frac{1}{2\lambda }\log \left(\frac{\lambda +n}{\lambda -v}\right)=\frac{g}{{\lambda }^2}t+c$

$\therefore \frac{1}{\lambda }\tan .\frac{1}{2}\log \left(\frac{1+v/\lambda }{1-v/\lambda }\right)=\frac{g}{{\lambda }^2}t+C$

Since $\tan h^{-1}x=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$

$\frac{1}{\lambda }\tan h^{-1}\frac{v}{\lambda }=\frac{g}{{\lambda }^2}t+c$

But by data, when $t=0,v=0$ $\therefore c=0$

$\therefore \frac{1}{\lambda }\tan h^{-1}\frac{v}{\lambda }=\frac{g}{{\lambda }^2}t$ $\therefore \tan h^{-1}\frac{v}{\lambda }=\frac{gt}{\lambda }$

$v=\lambda \tan h\left(\frac{gt}{\lambda }\right)$

But $v=\frac{dx}{dt}$, $\therefore \frac{dx}{dt}=\lambda \tan h\left(\frac{gt}{\lambda }\right)+C$. ...(1)

But by data, when $t=0,x=0$ $\therefore c=0$

$\therefore x=\frac{{\lambda }^2}{g}\log \cos h\left(\frac{gt}{\lambda }\right)$ ....(2)

Thus, the velocity of the body is given by (1) and the distance travelled is given by (2).