A train travels between two stations

[DeathEater]

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

 

[C. Lee]

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.

 

[haruspex]

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.

 

[DeathEater]

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!

 

[vela]

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.

 

[C. Lee]

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.