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 $$1.0$$ $$m/s2$$ and accelerates at a rate of $$0.50$$ $$m/s2$$ , how much time is lost in stopping at the station?

Question

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 $$1.0$$ $$m/s2$$ and accelerates at a rate of $$0.50$$ $$m/s2$$ , how much time is lost in stopping at the station?

Infinity Learn · Accepted Answer

Case $$-I$$: Actual time spent in $$deceleratingt=$$ $$v-u/a=20sDistance$$ travelled in $$decelerating=$$ $$20$$ × $$20-12$$×$$1$$×$$202$$ $$=$$ $$400-200$$ $$=$$ $$200m$$ Time that train would have taken if it had travelled uniformly with $$20$$ $$m/s$$  for $$200$$ m $$=$$ $$20020$$ $$=$$ $$10sExtra$$ time spent in decelerating $$=$$ $$20-10$$ $$=$$ $$10$$ $$sCase-II$$ : Actual time spent in $$acceleratingt=$$   $$v-ua$$ $$=$$ $$20-00.5$$ $$=$$ $$40$$ sDistance travelled in these $$40$$ $$s=$$ $$0$$×$$40+12$$×$$0.5$$×$$402$$ $$=$$ $$400$$ mTime that the train would have taken if it had travelled  uniformly with $$20$$ $$m/s$$ $$=$$ $$40020$$ $$=$$ $$20$$ sExtra time lost due to acceleration $$=$$ $$40-20$$ $$=$$ $$20$$ sTotal extra time $$=$$ $$10s$$ $$+$$ $$2$$ min $$+$$ $$20$$ s  $$=$$ $$2$$ min $$30$$ s

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