In the design of a rapid transit system , it is necessary to balance the average
ID: 1405528 • 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 9.0 km trip in two situations. (Assume that at each station the train accelerates at a rate of 1.1 m/s^2 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/s^2 . Assume it stops at each intermediate station for 20.0 s .)
Calculate the time it takes a train to make a 9.0 km trip if the stations at which the trains must stop are 1.5 km apart (a total of 7 stations, including those at the ends).
t1= ______s
Calculate the time it takes a train to make a 9.0 km trip if the stations are 3.0 kmapart (4 stations total).
t2=______s
Explanation / Answer
All calculations done in m and sec.
Total Distance = 9000 m
a = 1.1m/s^2
Maxmimum speed = 26.3889 m/s
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Time taken to reach max speed :
v = u + at
v = 0 + 1.1*t
26.3889 = 1.1*t
t = 23.9899 sec
Distance Travelled to reach max speed :
s = ut + 1/2*a*t^2
s = 0 + 1/2*1.1*(23.9899)^2
s = 316.533 m
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Time taken to de-accelerate from the max speed :
v = u + at
0 = 26.3889 - 2*t
t = 12.195 sec
Distance Travelled to stop
S = ut + 1/2*a*t^2
s = 26.3889*13.195 - (13.195)^2
s = 174.09 m
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1) Hence if stations are 1500 m apart then total 7 stations.
Then
Distance for which it runs at max speed is
1500 - 316.53 - 174.09 = 1009.37
Time taken to cover this distance is = 1009.37/26.3899 = 38.25 sec
Hence total time is 38.25 + 23.9899 + 13.195 = 75.43
Now
THe time calculated was for each stations. We have 6 stations.
Hence
total time = 75.43*6 +(waiting time)
= 75.43*6 + 20 * 5
= 552.60 sec
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2) Now if stations are 3000 m apart then total 4 stations.
Then
Distance for which it runs at max speed is
3000 - 316.53 - 174.09 = 2509.37
Time taken to cover this distance is = 2509.37/26.3899 = 95.09 sec
Hence total time is 95.09 + 23.9899 + 13.195 = 132.276
Now
THe time calculated was for each stations. We have 6 stations.
Hence
total time = 132.276*3 +(waiting time)
= 132.276*3 + 20 * 2
= 436.828 sec