A train travels between two stations

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In summary, the train travels between stations A and D, with uniform acceleration from A to B, constant velocity from B to C, and uniform deceleration from C to D. The distances AB, BC, and CD are all equal, and it takes 5 minutes to travel between the two stations. By dividing the graph of distance vs. time into separate shapes and using the equation 1/2(time of AB)*V= time for BC * V, it can be determined that the train spends 2 minutes traveling between A and B, 1 minute traveling between B and C, and 2 minutes traveling between C and D.
  • #1
DeathEater
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Homework Statement


a train travels between stations one and two (station one is marked by "A", and station two is marked by "D". There are also the other points "B" and "C") the engineer of the train is instructed to start from rest at station 1 and accelerate uniformly between points A and B, then coast with a uniform velocity between points B and C, and finally accelerate uniformly between points C and D until the train stops at station 2. The distances AB, BC, and CD are all equal, and it takes 5 minutes to travel between the two stations. Assume that the uniform accelerations have the same magnitude, even when they are in opposite directions.

a.) How much of this 5 minute period does the train spend traveling between points A and B?
b.)How much of this 5 minute period does the train spend traveling between points B and C?
c.) How much of this 5 minute period does the train spend traveling between points C and D?

Homework Equations


I have no idea which kinematic equations could be useful

The Attempt at a Solution


so I tried to make a list of "givens"

AB {Vi= 0 m/s}
BC {a=0 m/s^2}
CD {Vf= 0 m/s}
total time: 5 minutes
AB=BC=CD
Acceleration for AB= Acceleration for CD
 
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  • #2
Acceleration between AB : a
Acceleration between CD : -a
Time spent between AB : t_1
Time spent between BC : t_2
Time spent between CD : t_3
Total travel time is 5 min, so t_1+t_2+t_3 = 5 min

Distance between AB : 0.5*a*(t_1)^2 (1)
Distance between BC : a*t_1*t_2 (2)
Distance between CD : a*t_1*t_3 - 0.5*a*(t_3)^2 (3)

These three distances should be the same.
Equate these pairwisely to get the relations between t_1, t_2, and t_3

Now plug these into t_1+t_2+t_3 = 5 min, then you will get what you need.
 
  • #3
C. Lee said:
Acceleration between AB : a
Acceleration between CD : -a
Time spent between AB : t_1
Time spent between BC : t_2
Time spent between CD : t_3
Total travel time is 5 min, so t_1+t_2+t_3 = 5 min

Distance between AB : 0.5*a*(t_1)^2 (1)
Distance between BC : a*t_1*t_2 (2)
Distance between CD : a*t_1*t_3 - 0.5*a*(t_3)^2 (3)

These three distances should be the same.
Equate these pairwisely to get the relations between t_1, t_2, and t_3

Now plug these into t_1+t_2+t_3 = 5 min, then you will get what you need.
This is rather too much assistance at this stage. The system on these homework forums is to ask questions, provide hints and flag errors.

Necrophage, please post any equations you know which you suspect will be relevant. You mention kinematic equations, so list them.
 
  • #4
haruspex said:
This is rather too much assistance at this stage. The system on these homework forums is to ask questions, provide hints and flag errors.

Necrophage, please post any equations you know which you suspect will be relevant. You mention kinematic equations, so list them.
I'm sorry if this seems like I am trying to get anyone to do my work for me, I really just want to learn how to actually do this process. So I think I mostly used a graph that I drew to help me, and divided it into separate shapes to find the area of the graph. since the shape under the distance on a graph for AB is a triangle, I used 1/2(time of AB)*V= time for BC * V. I did it like that because they are equal distances but the shape for AB is a triangle and the shape under the graph for BC is a rectangle. I got rid of the V by dividing so I was left with the info that 1/2 time for AB is equal to the time for BC, and the same is true for the time taken to travel the distance for CD. So all in terms of BC, the time segments are 2BC+BC+2BC= 5 min. so the time taken for BC is exactly 1 min. And since I know that 1/2 time for AB is equal to the time for BC, then AB =2 minutes, and CD= 2 minutes.

so it follows: time for AB: 2 minutes +time for BC: 1 minute +time for CD:2 minutes= 5 minutes in total. Does that look correct? And also thanks for taking the time to reply to the original question!
 
  • #5
DeathEater said:
I'm sorry if this seems like I am trying to get anyone to do my work for me, I really just want to learn how to actually do this process.
Haruspex's comment was directed toward C. Lee, not you.

So I think I mostly used a graph that I drew to help me, and divided it into separate shapes to find the area of the graph. since the shape under the distance on a graph for AB is a triangle, I used 1/2(time of AB)*V= time for BC * V. I did it like that because they are equal distances but the shape for AB is a triangle and the shape under the graph for BC is a rectangle. I got rid of the V by dividing so I was left with the info that 1/2 time for AB is equal to the time for BC, and the same is true for the time taken to travel the distance for CD. So all in terms of BC, the time segments are 2BC+BC+2BC= 5 min. so the time taken for BC is exactly 1 min. And since I know that 1/2 time for AB is equal to the time for BC, then AB =2 minutes, and CD= 2 minutes.

so it follows: time for AB: 2 minutes +time for BC: 1 minute +time for CD:2 minutes= 5 minutes in total. Does that look correct? And also thanks for taking the time to reply to the original question!
Yes, this is correct, though I think you meant the plot was of velocity vs. time, not displacement vs. time. Good job.
 
  • #6
haruspex said:
This is rather too much assistance at this stage. The system on these homework forums is to ask questions, provide hints and flag errors.

Necrophage, please post any equations you know which you suspect will be relevant. You mention kinematic equations, so list them.

You are completely right. I'll be more careful from now on.
 

Related to A train travels between two stations

1. How fast does the train travel between the two stations?

The speed of the train between two stations can vary depending on various factors such as the type of train, the distance between the two stations, and any potential delays. Generally, trains can travel at speeds of up to 150 miles per hour, but this can also be affected by the terrain and weather conditions.

2. How long does it take for the train to travel between the two stations?

The duration of the train journey between two stations can also vary depending on the factors mentioned above. On average, a train can travel at a speed of 50 miles per hour, so a distance of 100 miles would take approximately 2 hours to cover. However, this can vary based on the stops and route the train takes.

3. How often do trains travel between the two stations?

The frequency of train travel between two stations can vary depending on the train schedule and the demand for that particular route. Generally, trains can run multiple times a day, with rush hour trains running more frequently than off-peak trains.

4. Can I purchase tickets for the train online?

Most train companies nowadays offer the option to purchase tickets online through their website or a third-party ticketing platform. This can be a convenient way to secure your ticket and avoid long queues at the station.

5. Are there any amenities available on the train?

Many trains offer amenities such as bathrooms, food and beverage options, and Wi-Fi. However, the availability of these amenities can vary depending on the type of train and the duration of the journey. It is best to check with the train company beforehand to see what amenities are available on your specific train journey.

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