In the design of a rapid transit system , it is necessary to balance the average
ID: 2002336 • 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 7.0 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 20.0 s .)
Part A:
Calculate the time it takes a train to make a 7.0 km trip if the stations at which the trains must stop are 1.4 km apart (a total of 6 stations, including those at the ends).
Part B:
Calculate the time it takes a train to make a 7.0 km trip if the stations are 3.5 km apart (3 stations total).
Explanation / Answer
For the first situastion. 6 stops. Work with one stop, thefirst one from the starting point to the first stop.
Divide the motion from the starting point to the first stop into 3 parts:
1- From the rest position to the cruising speed
2- From the cruising speed to complete stop
3- Motion at constant speed, (cruising speed)
For each part of that motion we need to find the time required and then the total time from rest to first stop.
Those three times we must calculate are t1, t2, and t3
In the process of solving for those 3 times , we have to find some distances too.
Start by converting the cruising speed from km/hr to m/s 95 km/hr x 1000m/km x 1 hr/3600 s = 26.4 m/s
Time to accelerate from zero to 26.4 m/s is given by the equation V= Vo +at1 where Vo (Initial velocity) = 0
and V is the cruising speed = 26.4 26.4 = 0 + (1.1) t1 and t1= 24sec
Next solve for t2 , the time to decelrate from the cruising speed of 26.4 to 0 called t2.
The same equation is used except the accel is deceleration and the sign of a is minus
V =V0 - at 2 Now V is the final velocity and = 0 and Vo is the initial velocity =26.4 (the cruising speed) and a = -2
0 = 26.4 - 2 t2 and t2 = 13.2s
In order to find the time t3 that the train travelled at cruising speed, we need to go back and calculate the distance covered during the acceleration and deceleration phases, t1 and t2, so we can subtract the sum of those two distances from the total distance for the first stop (1.2 km or 1200m) to use the balance of that distance to find t3
S1= Vo t1 +1/2 a(1)^2 where S1 is the distance covered during time t1 ,
Vo = 0 a= +1.1 = 0 + 0.5 (1.1)(24)^2 = 316 m
S2 = Vo t2 - 1/2 a t2^2 Vo is the cruising speed = 26.4 a= -2m/s^2 and t2 = 13.2 s = (26.4) (13.2) - 0.5 (2) (13.2) = 174m
Distance at constant(Cruising) speed = S3 = 1200 - 316 - 174 = 710 m
Time t3 at the cruising speed = Distance /velocity = 710/26.4 = 26.4
Total Time for one stop = t1 + t2 + t3 = 24 + 13.2 + 26.4 = 63,6 s
Total time from start to finish = 63,6 x 5 + 26 x4 = 421.6 sec the 26 sec is the iime at each stop.
1) Now the case of one stop at 3.5 km involves similar analysis.
From starting point to first stop. there is t1, t2 and t3 and also S1. S2 and S3. The;y are the same as the previous analysis except for t3 and S3. The other two are the same because the input for the rate of acceleraion and deceleration are the same.
2) For the second situration, the distance between the starti;ng point and the mid stop is 3500 m
The distancae at cruising speed is 3500 -S1 - S2 = 3500- 316 - 174 = 3010 m
From that we get t3 = 3010/26.4 = 114.1 sec
Total time for the second situation = 24 + 13.2 + 114.1 = 151.3 s from the rest position to the midpoint stop
Total time from start to finish = 151.3 x 2 + 26 x 1 = 328 .6 sec
There you have it 328.6 sec vs 421.6 sec or 5.48 min vs 7.03 min
26.4 = 0 + 1.1 t1
t1 =24 sec