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A train starts from rest from a station with acceleration $0.2 \,m / s ^2$ on a straight track and then comes to rest after attaining maximum speed on another station due to retardation $0.4 \,m / s ^2$. If total time spent is half an hour, then distance between two stations is [Neglect length of train]
$216$
$512$
$728$
$1296$
Solution
(a)
Shortcut : $S=\frac{1}{2} \frac{\alpha \beta}{\alpha+\beta} T^2$
$\alpha \rightarrow$ Acceleration
$\beta \rightarrow$ Deceleration (magnitude only)
$T \rightarrow$ Time of journey
$S \rightarrow$ Distance travelled
Given, $\alpha=0.2 \,ms ^{-2}$
$\beta=0.4 \,ms ^{-2}$
$T=$ half an hour $=30 \times 60 \,s =1800 \,s$
$S=\frac{1}{2} \times\left(\frac{0.2 \times 0.4}{0.2+0.4}\right) \times(1800)^2$
$\Rightarrow S=216000 \,m$
$\Rightarrow S=216 \,km$
