In the design of a rapid transit system, it is necessary to...

In the design of a $\textbf{rapid transit system}$, 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 15.0-km trip in two situations: ($a$) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and ($b$) the stations are 5.0 km apart (4 stations total). 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 22 s.

Key Concepts

Uniform Acceleration and Deceleration

This concept involves using kinematic equations to analyze motion when an object speeds up or slows down at a constant rate. It is essential for determining the time and distance required during the acceleration phase (to reach a cruise speed) and the deceleration phase (to come to a stop), which are necessary components in calculating overall journey times in transit systems.

Constant Velocity Motion

Once the train reaches its maximum speed, it travels at a constant velocity until it needs to slow down for the next station. The relationship between distance, speed, and time under constant velocity (distance = speed × time) is crucial for analyzing the cruising portions of the trip where acceleration is not a factor.

Average Speed Calculation

Average speed is found by dividing the total trip distance by the total travel time, including periods of acceleration, constant speed, deceleration, and dwell times at stations. This concept is important in transit design for understanding the impact of stop frequency on the effective speed of the system.

Dwell Time

Dwell time refers to the period the train spends at each station, not in motion, while handling passenger boarding and alighting. It directly contributes to the total travel time and is a critical factor in optimizing schedules and overall system performance.

Transit System Design Trade-offs

In rapid transit systems, a key design consideration is the balance between station spacing and overall journey time. Closer stations yield more frequent stops, leading to increased acceleration and deceleration phases and additional dwell time, thereby reducing average speed. Conversely, wider spacing can improve run speeds but may reduce accessibility. Understanding these trade-offs is essential for efficient transit planning.

Transcript

00:01 For this problem, on the topic of kinematics in one dimension, we are looking at the process of designing a rapid transit system.

00:09 And we want to look at two situations.

00:11 A train needs to make a 15 -kilometer trip, firstly, where the stations at which the train must stop are three kilometers apart, which gives a total of six stations.

00:23 And the second situation are where these stations are five kilometers apart, with four stations in total.

00:29 If we have the acceleration of the train, the maximum speed that it can reach, and then the deceleration, and the time it stops at each station, we want to calculate the time it will take the train to make this trip in both situations.

00:48 So firstly, we can analyze this problem as a series of three one -dimensional motions, the acceleration phase, the constant speed phase, and the deceleration phase.

00:58 Now, we can first find the maximum speed of the train, which will always be the same.

01:05 So the maximum speed is given in kilometres per hour, and this is 95 kilometers per hour, and we'll convert this to meters per second.

01:17 So that's one meter per second over 3 .6 kilometers per hour.

01:27 So converting kilometers to meters and hours to seconds.

01:30 This gives us the maximum speed of a train in si units to be 26 .39 meters per second.

01:42 Now in the acceleration phase the initial velocity v0 is equal to 0, and the acceleration itself is 1 .1 meters per square second.

01:50 Now we also know the final velocity.

01:53 So we can find the lapse time for the acceleration phase.

02:00 So for the acceleration of the train, we have the final velocity, velocity v is equal to the initial velocity v0 plus a times t so if we rearrange this equation we can solve for the time it takes the train to accelerate to maximum speed so t is v minus v v0 over a and this is 26 .39 meters per second minus 0 divided by 1 .1 meters per second squared.

02:41 And so if we calculate, we get the time for the train to accelerate to maximum speed to be 23 .99 seconds.

02:52 Now we want to find the displacement during the acceleration phase.

02:57 So the displacement is x minus x0.

03:02 And this is for the acceleration phase, so subscript ac c.

03:06 We know that this is equal to v0 t using another kinematic equation plus a half a t squared, where the time here is the time taken for the acceleration.

03:19 So that's zero, since the initial velocity is zero, plus a half times the acceleration of 1 .1 meters per square second times the time of 23 .99 seconds all squared, which gives us the displacement during this phase to be 3 .9 .5.

03:42 316 .5 meters.

03:46 Now in the deceleration phase we know the initial velocity v0 to be 26 .39 meters per second.

03:58 So for the deceleration phase we know the initial velocity and we know the deceleration which is minus 2 meters per square second.

04:09 We also know that the train will finally come to rest so its final velocity is 0.

04:13 So using the same equation we can find the time that the train takes to decelerate to rest.

04:18 So v is equal to v0 plus a t, which means that the time taken for deceleration t is equal to v minus v0 over a.

04:34 That's 0 minus 26 .39 meters per second divided by the acceleration of minus 2 meters per second, which gives us the deceleration of minus 2 meters per square second, which gives us the deceleration.

04:50 Acceleration time of 13 .2 seconds.

04:56 So now the distance travel during deceleration is x minus x0 x0 subscript d e c for deceleration and we use the equation again that we used above for the acceleration v0 t plus a half a t squared and so that's the initial velocity of 26 .39 meters per second times a time of 13 .2 seconds plus a half into minus two meters per square second, which is deceleration times t squared, which is 13 .2 squared.

05:49 And that's seconds...

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