In the design of a rapid transit system , it is necessary to balance the average
ID: 1637932 • 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 15 km trip in two situations:
Part A
the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends). Assume that at each station the train accelerates at a rate of 1.1 m/s2 until it reaches 94 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 25 s .
Express your answer to two significant figures and include the appropriate units.
Part B
the stations are 5.0 km apart (4 stations total).
Express your answer to two significant figures and include the appropriate units.
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 15 km trip in two situations:
Part A
the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends). Assume that at each station the train accelerates at a rate of 1.1 m/s2 until it reaches 94 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 25 s .
Express your answer to two significant figures and include the appropriate units.
Part B
the stations are 5.0 km apart (4 stations total).
Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
94km/h * 1m/s / 3.6km/h = 26.1 m/s
time to accelerate t' = 26.1m/s / 1.1m/s² = 23.74 seconds
time to decelerate t" = 26.1m/s / 2m/s² = 13.05 seconds
While accelerating and decelerating the average speed is
Vavg = 26.1m/s / 2
s = v*t
15000 m = 26.1m/s * (23.74/2 + t + 13.05/2)s for t, cruise time, in seconds
t = 556.31 s
for a total time between 2 stations of
T = (556.31 + 23.74 + 13.05)s = 593.1 seconds
A) For six stations: each station-station trip is such that
3000 m = 26.1m/s * (23.74/2 + t + 13.05/2)s
t = 96.5 s, and
T = 5 * (23.74 + 96.5 + 13.05) s + 4*22s = 666.45s + 88s = 754.5 seconds
B) 5000 m = 26.1m/s * (23.74/2 + t + 13.05/2)s
t = 173.2 s, and
T = 4 * (23.74 + 173.2 + 13.05)s + 3*22s = 840s + 66s = 906 seconds
Hope this helps!