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Q. 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]

Motion in a Straight Line

Solution:

$S=\frac{1}{2} \frac{\alpha \beta}{\alpha+\beta} T^{2}$
$\alpha \rightarrow$ 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 $=30 \times 60\, s =1800\, s$
$\Rightarrow S=\frac{1}{2} \times\left(\frac{0.2 \times 0.4}{0.2+0.4}\right) \times(1800)^{2}$
$\Rightarrow S=216000\, m$
$\Rightarrow S=216\, km$