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 stationalong 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=v−ua=0−20−1.0=20sDistance travelled in decelerating=20×20−12×1×202=400−200=200mTime that train will have taken if it had travelled uniformlywith 20 m/s for 200 m =20020=10sExtra time spent in decelerating = 20 - 10 = 10 sCase-II: Actual time spent in accelerating=v−ua=20−00.5=40sDistance travelled in these 40 s=0×40+12×0.5×402=400mTime that the train will have taken if travelled uniformly with 20 m/s=40020=20s Extra time lost due to acceleration =40−20=20s Total extra time =10s+2min+20s=2min30s