Question

A train is moving on straight track with velocity $v_{0}=13.5\, ms ^{-1}$. To stop the train at a particular station, the driver applies brakes at $t=0$, which is caused of a retardation proportional to the velocity of the train. The speed of train reduces $50 \%$ in the first $t_{0}=4$ in $2\, s$. The velocity of train (in $ms ^{-1}$ ) at $t=4 s$ (Given, $e=2.7$ ) is_______.

Answer 1

$\because a=-k v$ or $\frac{d v}{d t}=-k v$
Or $ \int\limits_{v_{0}}^{v} \frac{d v}{v}=-k \int\limits_{0}^{t} d t$
or $[\ln v]_{v_{0}}^{v}=-k t$
Or In $v-\ln v_{0}=-k t$
or $\ln \frac{v}{v_{0}}=-k t$
Or $\frac{v}{v_{0}}=e^{-k t}$
For $t=t_{0}, v=v_{0}-v_{0} \times \frac{50}{100}=\frac{v_{0}}{2}$
$\therefore \frac{v_{0}}{2}=v_{0} e^{-k t}$
Or $v=v_{0} e^{\frac{-\ln 2}{t_{0}} t}$
or $\ln v=\ln v_{0}-\frac{\ln 2}{t_{0}} t$
Or $\ln \frac{v}{v_{0}}=-\frac{\ln 2}{4 \ln 2} t$
or $\ln \frac{v}{v_{0}}=-\frac{t}{4}$
or $\frac{v}{v_{0}}=e^{-\frac{t}{4}}$
$\therefore v=v_{0} e^{-\frac{t}{4}}=13.5 e^{-\frac{4}{4}}$
$=\frac{13.5}{e}=\frac{13.5}{2.7}=5\, ms ^{-1}$