The Train Puzzle
 

This one might be too easy for a bunch of Math-ers.
========================================
Two men are walking toward each other along a railway track
At time 0 a freight train meets up with the first man.
It overtakes the first man in 20 seconds.
10 minutes later it meets the second man coming the other way.
The train passes the second man in 18 seconds.
How long after the train has passed him will the two men meet??
Assume constant speeds throughout.
However, do NOT assume both men are walking the same speed.

[spoiler]
L is length of train
velocity of train relative to first man is L/20
Velocity of second man relative to train is -L/18
velocity of second man relative to first man is -L/18 + L/20 = -L/180
When train reaches second man, their separation is 10mins*L/20
Time at which they meet is 90 minutes later
[/spoiler]

David

[quote=davieddy;102551][spoiler]
L is length of train
velocity of train relative to first man is L/20
Velocity of second man relative to train is -L/18
velocity of second man relative to first man is -L/18 + L/20 = -L/180
When train reaches second man, their separation is 10mins*L/20
Time at which they meet is 90 minutes later
[/spoiler][/quote]Suppose we assign specific constants to the first man's speed and the train's length:

(1) the first man is traveling south at 1 meter/second.

(2) the train is 1000 m long.

[spoiler]and assign variable x to the second man's speed:

(3) the second man is traveling north at x m/s.

During the 20 seconds the train is overtaking the first man, that man traveled 20 m south, so the distance traveled by the train in the same time is 1020 m. So, train is traveling south at 51 m/s.

During the 18 seconds it is passing the second man, the train traveled 18 * 51 = 918 m south, while the second man traveled north 18*x meters. So 1000 m = 918 m + 18 * x. So x = 82/18 = 41/9 = 4 + 5/9. The closure rate of the men is 5 + 5/9 = 50/9 m/s ( = L/180 per second, in accordance with davideddy).

During the 10 minutes between when it finished overtaking the first man and when it met the second man, the train traveled 600 * 51 = 30600 m/s south and the second man traveled 41/9 * 600 = 8200/3 m north. (Also, the first man traveled 600 m south.) So when the train just finished overtaking the first man, the distance between its front and the second man was 30600 m + 8200/3 m, and the distance between the two men was 30600 m + 8200/3 m + 1000 - 600 m = 33733 + 1/3 m.

10 minutes and 18 seconds later, the men were 618 * 50/9 = 30900/9 = 3433 + 1/3 m closer together.

33733 + 1/3 m - (3433 + 1/3) m = 30300 m.

30300 / (50/9) = 5454 seconds after the train finishes passing the second man, which does not depend on x.[/spoiler]

Then I changed the first man's speed to y m/s, but when I finished my answer was incompatible with [spoiler]the 5454 figure above[/spoiler] when y = 1, so there's some error in my calculations somewhere.

You are interpreting "ten minutes later" as the time between
the rear of the train passing man 1 and the front passing man 2.
I was referring all times to the front of the train passing the men.
(Including my answer).
I'm still trying to spot your error. (Marking work like this is what
made me stop teaching:smile: )

[spoiler]
Relative to man 1, the train travels at L/20 and man2 at L/180.
If it takes 618s between the rear of the train passing man1 and man2,
their separation is then 618s * L/20.
The time taken for them to meet is 9*618s = 5562s
[/spoiler]

David

[QUOTE=cheesehead;102573]So when the train just finished overtaking the first man, the distance between its front and the second man was 30600 m + 8200/3 m, and the distance between the two men was 30600 m + 8200/3 m + 1000 - 600 m = 33733 + 1/3 m[/QUOTE]
The subtraction of 600m is superfluous.

[quote=axn1;102592]The subtraction of 600m is superfluous.[/quote]

Yes. If you leave it out, you get 5562s, same as I did when I interpreted
"10 minutes later" as you did.

You were computing the distance between the two men at
the start of the ten minutes, before man1 had moved 600m.

My approach is simpler because it treats man1 as stationary.
No loss of generality accrues because of the use of "relativity".
(Gallilean as opposed to Einsteinian I hasten to add)

Also the question of how far apart they were when the
train passed man1 was unasked for and a roundabout approach.

We have three objects to consider which is quite enough
without bringing the earth into it!

David

[QUOTE=davieddy;102610]Yes. If you leave it out, you get 5562s, same as I did when I interpreted
"10 minutes later" as you did.[/QUOTE]

Correct ... at least that is the same answer I got!!!!!!

[quote=davieddy;102582]You are interpreting "ten minutes later" as the time between the rear of the train passing man 1 and the front passing man 2.
I was referring all times to the front of the train passing the men.[/quote]Oh, okay. That's why I was trying specific numbers for speed and length.

[CODE]At time 0 a freight train meets up with the first man.
It overtakes the first man in 20 seconds.
10 minutes later it meets the second man coming the other way.
[/CODE]

I would interpretate this as the train meets the second man at time 10min 20sec.

Is the english language ambiguous about which sentence "later" referes to?

-Eivind

[quote=Eivind;102966]
I would interpretate this as the train meets the second man at time 10min 20sec.

Is the english language ambiguous about which sentence "later" referes to?

-Eivind[/quote]

That's how we interpreted it.
But having set up this way of allocating a time to each event,
the setter of the problem abandoned it.

I initially thought he meant that the train met the second man
at time 10 minutes, and meant to ask "at what time do the men meet?"

David

[quote=cheesehead;102941]Oh, okay. That's why I was trying specific numbers for speed and length.[/quote]

If the answer depended on these, they would be specified in the question.
It doesn't, (wherein lies the attractiveness of the problem).
But from your convoluted (mis)calculation, this may not be
immediately apparent:smile:

David

Let me be more specific....
 

Two men (A and B) are walking toward each other along a railway track as in the amazingly lifelike :smile: picture below.
A is walking EAST; B is walking WEST
(W) === A--> ============================= <--B === (E)
At time 0 a freight train meets up with the first man.
=== The train is also heading EAST.
=== At time 0 the FRONT of the train is even with man A
It overtakes the first man in 20 seconds.
=== At time 20 seconds the BACK of the train is even with man A
10 minutes later it meets the second man coming the other way.
=== At time 10:20 the FRONT is the train meets man B
The train passes the second man in 18 seconds.
=== At time 10:38 the END of the train leaves man B
How long after the train has passed him will the two men meet??
=== How long after time 10:38, that is
Assume constant speeds throughout.
However, do NOT assume both men are walking the same speed.

Yes. We all guessed that.

[quote=davieddy;102984]If the answer depended on these, they would be specified in the question.
It doesn't, (wherein lies the attractiveness of the problem).[/quote]What I meant was that I was trying specific numbers to see whether I was interpreting the problem wording correctly.