A train normally travels at a uniformly speed of 72 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 on this track. If the train decelerates at a uniform rate of 1.0 $\mathrm{m}/{\mathrm{s}}^2$ and accelerates at a rate of 0.50 $\mathrm{m}/{\mathrm{s}}^2$ , how much time is lost in stopping at the station?  

Case -I: Actual time spent in decelerating

t= v-u/a=20s

Distance travelled in decelerating

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

Time that train would 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

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

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 would have taken if it had travelled  uniformly with 20 m/s = $\frac{400}{20} = 20 \mathrm{s}$

Extra time lost due to acceleration = 40-20 = 20 s

Total extra time = 10s + 2 min + 20 s  = 2 min 30 s