Two trains leave a station, and an observer misses them both what does he see

[stunner5000pt]

Using Speical relativity, of course

If two trains leave a station on the same track. An observer missed both these trains and is standing close to the track sees the westbound train recede at 0.6c and sees the eastbound train recede at 0.8c. There is a ticket collector on the westbound train going from the back of the train to the front at 0.4c, with respect to a passenger on the westbound train.

What would the speed of the eastbound train wit hrespect to the westbound train (call it Ur) according to:

a) Observer on the station: it's just as if the trains were approaching each other... right?

in taht case using Ux' = UW - UE / (1 - UW UE / C^2) yields 0.38c

b)Passenger seated on the westbound train?

The same framework as the previous question (which leads me to doubt part A) and i get 0.38c

c) Ticket collector on the westbound train?
The ticket collector would see hte same as the passenger, no?

Similarly what is the speed of the ticket collector :

d) According to the observer on the station?
I would think calculating the speed of the ticket collector with respect tothe train first, and then the observer to the train and then adding the velocities up

e) According to the passenger on the east train?
First find the relative velocities of the two trains and then add the velocity of the ticket collector with respect to the west train as done in the previous one

F) Relative to a passenger seated on the east train according tothe observer on the station?

STumped even more...

Please please help I'm desperate for help!

i would appreciate your help

 

[Doc Al]

What would the speed of the eastbound train wit hrespect to the westbound train (call it Ur) according to:

a) Observer on the station: it's just as if the trains were approaching each other... right?

in taht case using Ux' = UW - UE / (1 - UW UE / C^2) yields 0.38c

Does that even make sense? After all, the speed of the eastbound train with respect to the platform is 0.8c--how can its speed with respect to the westbound train be less? :yuck:

Learn the relativistic velocity addition formula:
$$V_{c/a} = \frac{V_{b/a} + V_{c/b}}{1 + V_{b/a}V_{c/b}/c^2}$$
(Hint: signs--which signify direction--matter!)

 

[wais]

in this matter!

According to special relativity, the observer on the station would see the two trains approaching each other at a relative speed of 0.38c. This is because time and space are perceived differently for objects in motion, making it appear as though the trains are moving at different speeds when in reality they are moving at the same speed relative to the observer on the station.

For the passenger seated on the westbound train, they would also see the trains approaching each other at a relative speed of 0.38c. This is because their perception of time and space is also affected by their motion on the train, leading to the same result as the observer on the station.

The ticket collector on the westbound train would see the same as the passenger, as they are both moving at the same speed on the train.

To calculate the speed of the ticket collector according to the observer on the station, you would first need to find the relative velocity of the ticket collector with respect to the train. This would be 0.4c, as stated in the question. Then, you would need to add this velocity to the velocity of the train with respect to the observer on the station. This would give you a final speed of 0.78c.

Similarly, for the speed of the ticket collector according to the passenger on the eastbound train, you would first find the relative velocities of the two trains, which is 0.2c. Then, you would add the velocity of the ticket collector with respect to the westbound train, which is 0.4c. This would give you a final speed of 0.6c.

For the relative speed of the passenger on the eastbound train according to the observer on the station, you would use the same approach as in part A. You would find the relative velocities of the two trains, which is 0.2c, and then add this to the velocity of the ticket collector with respect to the westbound train, which is 0.4c. This would give you a final speed of 0.38c.

I hope this helps clarify the situation for you. Remember, special relativity can be a complex concept to understand, but with practice and understanding of the principles, it can become easier to work with. Don't hesitate to seek further help or clarification if needed.