In the design of a rapid transit system, it is necessary to balance the average
ID: 2002922 • 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 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 6.0 km trip in two situations. (Assume that at each station the train accelerates at a rate of 1.1 m/s2 until it reaches 95 km/h , then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at -2.0 m/s2 . Assume it stops at each intermediate station for 19.0 s .) 1) Calculate the time it takes a train to make a 6.0 km trip if the stations at which the trains must stop are 1.2 km apart (a total of 6 stations, including those at the ends). 2) Calculate the time it takes a train to make a 6.0 km trip if the stations are 3.0 km apart (3 stations total).
Explanation / Answer
a)Total Distance covered= 6 km = 6000 m
Total number of Stations = 6
distance between stations = 1.2 km = 1200 m
max speed = 95 km/hr = 95 km/hr x ( 1000m/1km) x (1hr/3600s) = 26.4 m/s
acceleration = 1.1 m/s2
deceleration = -2 m/s2
Time to accelerate from rest = t1
t1= (v-u)/a = (26.4 - 0)/1.1 = 24 s
Distance to reach max speed of 95km/hr
s1= (v2 - u2)/2a = (26.4)2/2(1.1) = 316.8 m
Time to decelerate before stopping at next station = t3
t3=(v -u)/a = (0 - 26.4)/-2 = 13.2 s
Distance to reach max speed of 95km/hr
s3= (v2 - u2)/2a = 0-(26.4)2/2(-2) =177.24 m
Distance travelled with constant velocity s2= 1200 - (s1+s3) = 1200 - (316.8 +177.24) =708.96 m
Time travelled with constant velocity t2= 708.96 / 26.4 =26.85 s
Total travel time between stations = 24 +13.2 + 26.85 = 64.05 s
Total travel time = travel time between 5 stations + waiting time for 4 stations = 5(64.05) + 4(19) = 396.25 s = 6.6min
b)Total Distance = 6 km = 6000 m
Stations = 3
distance between stations = 3 km = 3000 m
Distance travelled with constant velocity s2= 3000 - (s1+s3) = 3000 - (316.8 +177.24) = 2505.96 m
Time travelled with constant velocity t2= 2505.96/ 26.4 = 94.92 s
Total travel time between stations = 24 + 94.92+ 26.85 = 145.77 s
Total travel time = travel time between 3 stations + waiting time for 2 stations = 3(145.77) + 2(19) = 475s = 7.92 min