In the design of a rapid transit system, it is necessary to balance the average
ID: 1361131 • Letter: I
Question
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between station stops. The more stops there are, the slower the train's average speed. To get an idea of this problem, calculate the time it takes a train to make a 31.0-km trip in two situations. In each case, assume that at each station the train accelerates at a rate of 1.30 m/s2 until it reaches a speed of 92.0 km/h, then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at -2.30 m/s2. Assume also that the train stops at each intermediate station (those not at the ends) for 20. s.
How long does it take the train to make the trip if there are a total of 6 stations (including those at the ends), which are 6.20 km apart? Give your answer in minutes.
How long will it take if there are a total of 4 stations, which are 10.3 km apart?
Explanation / Answer
Given that,
v(avg) = 92 km/h = 25.56 m/s ; a1 = 1.3 m/s2 ; a2 = -2.3 m/s2 ; D = 31 km = 31000 m
a = v/t => t1 = v/a = 25.56/1.3 = 19.66 s
t2 = v/a2 = 25.56/-2.3 = 11.11 sec
We need to know the cruise time as well, let it be t
distance = speed x time
31000 = 25.56 x (19.66/2 + t + 11.11/2)
393.24 + 25.56 t = 31000
t = 1197.4 sec
Total time between two stations:
T = 1197.4 + 19.66 + 11.11 = 1228.17 sec
D' = 6.2 km = 6200 m ; n1 = 6
Vavg x total time = 6200
25.56 x (19.66/2 + t + 11.11/2) = 6200
393.24 + 25.56 t = 6200
t = 277.2 sec
total time the train takes to make a trip, if there are 6 stations
T(6) = n1 x T
T = 227.2 + 19.66 + 11.11 = 257.97
T(6) = 6 x 257.97 = 1547.82 sec = 25.8 minutes
Hence, T(6) = 25.8 minutes
Now when ; D'' = 10.3km = 10300 m ; n2 = 4
25.56 x (19.66/2 + t + 11.11/2) = 10300
t = 387.6
T = 19.66 + 387.6 + 11.11 = 418.4
T(4) = 4 x 418.4 = 1673.6 sec = 27.9 minutes
Hence, T(4) = 27.9 minutes