In the design of a rapid transit system, it is necessary to balance out the aver
ID: 2043425 • Letter: I
Question
In the design of a rapid transit system, it is necessary to balance out the averagespeed of a train against the distance between stops. The more stops there are , the
slower the average speed. Calculate the time it takes for a train for a30 km trip in
2 situations (a) the stations at which the train must stop are 1 km apart (b) te
stations are 3 km apart. Assume at each station the train accelerates at 1.5 m/s2,
until it reaches 80 Km/h, stays at that speed until it applies brakes for deceleration
at -3 m/s2 to arrive at the next station to stop for 20 s. From (a) and (b) above
write a general formula using the parameters for the average speed.
Explanation / Answer
case (a) each station is 1 km apart
Total stop at 29 stations = 29x20 = 580s
Time travel between station = taccelerate + tconstant + tdecerelate
80km/h = 22.2m/s
taccelerate = (22.2-0) / 1.5 = 14.8s
distance during this period = 0.5*(1.5)*(14.82) =164.28m
tdecelerate = (0-22.2) / (-3) = 7.4S
distance during this period = 0.5*(3)*(7.42) =82.14m
distance left for constant speed period = 1000-164.28-82.14 = 753.58m
tconstant = 753.58/22.2 = 33.95s
Time travel between station = 14.8+33.95+7.4 = 56.15s
Total time fo the trip = 580 + 56.15*29 = 2208.35s
Average speed = Total distance / Total time = 30000 / 2208.35 = 13.585m/s = 48.91km/h
case (b) each station is 3 km apart
Total stop at 9 stations = 9x20 = 180s
Time travel between station = taccelerate + tconstant + tdecerelate
80km/h = 22.2m/s
taccelerate = (22.2-0) / 1.5 = 14.8s
distance during this period = 0.5*(1.5)*(14.82) =164.28m
tdecelerate = (0-22.2) / (-3) = 7.4S
distance during this period = 0.5*(3)*(7.42) =82.14m
distance left for constant speed period = 3000-164.28-82.14 = 2753.58m
tconstant = 2753.58/22.2 = 124.04s
Time travel between station = 14.8+124.04+7.4 = 146.24s
Total time fo the trip = 180 + 146.24*9 = 1496.16s
Average speed = Total distance / Total time = 30000 / 1496.16 = 20.051m/s = 72.18km/h