An open car of mass $m_{0}$ is running on smooth horizontal rails under rain falling vertically which it catches and retains in it. If $v_{0}$ is the initial velocity of the car and $\mu$ the mass of water falling into the car per second, find the distanec travelled by it in time $t$.
Solution
$m \frac{d v}{d t} \mid v \frac{d i n}{d t}-0 \quad$ So, $\left(m_{0}+\mu t\right) \frac{d v}{d t} \mid \mu v-0 \quad \therefore \quad \frac{d v}{v}-\frac{-\mu d t}{m_{o} \mid \mu t}$
Integrating, $\ln v=-\ln \left(m_{0}+\mu i\right)+C \quad$ So, $C=\ln v_{0}+\ln m_{0} \quad$ because when $t=0, v=\nu_{0}$
$\therefore \quad \ln \frac{v}{v_{0}}--\ln \left(m_{0}+\mu t\right)+\ln m_{0}-\ln \frac{m_{a}}{m_{o}+\mu t}$
$\therefore \quad v=v_{s} \frac{m_{\theta}}{m_{o}+\mu t}=\left(\frac{d s}{d t}\right) \quad \therefore \frac{d t}{m_{o}+\mu t}=\frac{d s}{m_{o} v_{0}} ;$ Integrating, we get
$\therefore \quad \frac{1}{\mu} \ln \left(m_{o}+\mu \theta\right)=\frac{s}{m_{0} v_{0}}+D$
when $t=0, s=0, \quad \therefore \quad \frac{1}{\mu} \ln m_{o}=D \quad \therefore \quad \frac{1}{\mu} \ln \left(m_{o}+\mu t\right)=\frac{S}{m_{o} v_{o}}+\frac{1}{\mu} \ln m_{o}$
$\Rightarrow \frac{1}{\mu} \ln \frac{\left(m_{0} \mid \mu t\right)}{m_{y}}-\frac{s}{m_{o} v_{o}} \quad \therefore \quad s-\frac{m_{0} v_{\theta}}{\mu} \ln \left(1+\frac{\mu t}{m_{0}}\right)$