Question

The differential equation of a body projected from the earth is given by $\frac{dv}{dt}=-g-kv$.The distance travelled by the body at any time $t$ is given by $x=(\frac{g}{k^2}+\frac{u}{k})(1+e^{-kt})-\frac{g}{k}t$

• A. True
• B. False

Answer 1

The correct option is B False

We have $\frac{dv}{dt}=-g-kv$. $\therefore \frac{dv}{g+kv}=-dt$

By integration, we get $\frac{1}{k}\log (g+kv)=-t+c$

Initially when $t=0,v=u$. $\therefore \frac{1}{k}\log (g+ku)=c$

$\therefore \frac{1}{k}\log (g+kv)=-t+\frac{1}{k}\log (g+ku)$

$\therefore t=\frac{1}{k}\log \left(\frac{g+ku}{g+kv}\right)$ $\therefore \frac{g+ku}{g+kv}=e^{kt}$

$\therefore (g+ku)e^{-kt}=g+kv$

$\therefore v=-\frac{g}{k}+\left(\frac{g+ku}{k}\right)e^{-kt}$ $\therefore \frac{dx}{dt}=-\frac{g}{k}+\left(\frac{g+ku}{k}\right)e^{-kt}$

By integration, we get $x=-\frac{g}{k}t-\left(\frac{g+ku}{k^2}\right)e^{-kt}+c$

initially when $t=0,x=0$ $\therefore c=\frac{g+ku}{k^2}$

$\therefore x=-\frac{g}{k}t-\left(\frac{g+ku}{k^2}\right)e^{-kt}+\left(\frac{g+ku}{k^2}\right)$

$=\left(\frac{g+ku}{k^2}\right)(1-e^{-kt})-\frac{g}{k}t$

$=(\frac{g}{k^2}+\frac{u}{k})(1-e^{-kt})-\frac{g}{k}t$.