In the design of a rapid transit system , it is necessary to balance the average
ID: 1527600 • 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 8.0 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 27.0 s .)
A.Calculate the time it takes a train to make a 8.0 km trip if the stations at which the trains must stop are 2.0 kmapart (a total of 5 stations, including those at the ends).
Express your answer using two significant figures.
B.Calculate the time it takes a train to make a 8.0 km trip if the stations are 4.0 km apart (3 stations total).
Express your answer using two significant figures.
Explanation / Answer
95km/h * 1m/s / 3.6km/h = 26.4 m/s
time to accelerate t' = 26.4m/s / 1.1m/s² = 24 seconds
time to decelerate t" = 26.4m/s / 2m/s² = 13.2 seconds
While accelerating and decelerating the average speed is
Vavg = 26.4m/s / 2
s = v*t
8000 m = 26.4m/s * (24/2 + t + 13.2/2)s for t, cruise time, in seconds
t = 284.56 s
for a total time between 2 stations of
T = (284.56 + 24 + 13.2)s = 315 seconds
A) For six stations: each station-station trip is such that
2000 m = 26.4m/s * (24/2 + t + 13.2/2)s
t = 57.2 s, and
T = 4 * (24 + 57.2 + 13.2)s + 3*27s = 377.6s + 81s = 458.6 seconds
B) 4000 m = 26.4m/s * (24/2 + t + 13.2/2)s
t = 132.9 s, and
T = 2 * (24 + 132.9 + 13.2)s + 1*27s = 340.2s + 27s = 367.2 seconds