train normally travels at a uniform speed of 72V,km/h on a long stretch of straight level track. On a particular day, the train was forced to make a 2.0 minute stop at a station
along this track. If the train decelerates at a uniform rate of 1.0 m/s² and accelerates at a rate of 0.50 m/s², how much time is lost in stopping at the station?

Case-I: Actual time spent in decelerating

$=\frac{v-u}{a}=\frac{0-20}{-1.0}=20\mathrm{s}$

Distance travelled in decelerating

$=20\times 20-\frac{1}{2}\times 1\times {20}^2=400-200=200\mathrm{m}$

Time that train will have taken if it had travelled uniformly

with 20 m/s for 200 m $=\frac{200}{20}=10\mathrm{s}$

Extra time spent in decelerating = 20 - 10 = 10 s

Case-II: Actual time spent in accelerating

$=\frac{v-u}{a}=\frac{20-0}{0.5}=40\mathrm{s}$

Distance travelled in these 40 s

$=0\times 40+\frac{1}{2}\times 0.5\times {40}^2=400\mathrm{m}$

Time that the train will have taken if travelled uniformly with 20 m/s

$=\frac{400}{20}=20\mathrm{s}$

$\begin{array}{l}\text{ Extra time lost due to acceleration }=40-20=20\mathrm{s}\\ \text{ Total extra time }=10\mathrm{s}+2\min +20\mathrm{s}=2min30\mathrm{s}\end{array}$