Question

A train normally travels at a uniform speed of 74km/h on a long stretch of straight, level track. On a particular day, the train must make a 3.0min stop at a station along this track.

If the train decelerates at a uniform rate of 1.3m/s2 and, after the stop, accelerates at a rate of 0.70m/s2 , how much time is lost because of stopping at the station?

Answer 1

Vo = 74km/h = 74000m/3600s = 20.56 m/s.

V = Vo + a*t
t1 = (V-Vo)/a = (0-20.56)/-1.3 = 15.8 s.

t2 = 3 min = 180 s.

t3 = 20.56-0)/0.7 = 29.37 s

T=t1 + t2 + t3=15.8 + 180 + 29.37=225.2s
= 3.75 min. = Time lost.