One Train, Two Train, Red Train ,Blue Train.

Joe is at the rail station when he notices two trains approaching each other from opposite directions on parallel tracks. One is red an one is blue. The two trains begin to cross paths directly in front of Joe ( the front of the red train is coincident with the front of the blue train ).

The trains move at a constant speed ( meters per second ):

v_blue = 20 m/s
v_red = 30 m/s

Joe's Observations:

The time it takes for the trains to cross each others path ( the end of the blue train is coincident with the end of the red train ) is 20 seconds.

and

The end of the blue train reaches Joe's position 10 seconds after the end of the red train reached his position, how long is the red train?

OK, so the trick here is to realise that the time it takes for one train at 20 m/s to overtake another train at -30 m/s is the same as the time it takes one train at 50 m/s to overtake a stationary one. (Or a stationery one. Who knows? I'm not going to stop you from origami'ing Mallard out of A4 paper.)

If you draw a diagram, you'll see that this means that if it takes 20 s at a combined 50 m/s for the trains to pass, they have to be 20 * 50 = 1000 m in total.

Now, the end of the red train reaches Joe after l_red/30 s, and the end of the blue train reaches him after l_blue/20 s; so l_red/30 = l_blue/20 + 10.

We now have two equations and two variables:
20 l_red = 30 l_blue + 300
l_red + l_blue = 1000 <=> 20 l_red + 20 l_blue = 20000 <=> 20 l_red = 20000 - 20 l_blue

Thus, 30 l_blue + 300 = 20000 - 20 l_blue <=> 50 l_blue = 19700 <=> l_blue = 19700/50 = 394 m.

So the blue train is 394 meters long, and the red one is 606 meters.

@shallow-blue said
OK, so the trick here is to realise that the time it takes for one train at 20 m/s to overtake another train at -30 m/s is the same as the time it takes one train at 50 m/s to overtake a stationary one. (Or a stationery one. Who knows? I'm not going to stop you from origami'ing Mallard out of A4 paper.)

If you draw a diagram, you'll see that this means that if it take ...[text shortened]... > l_blue = 19700/50 = 394 m.

So the blue train is 394 meters long, and the red one is 606 meters.

Sorry, incorrect ( but close ).

Examine this derivation more carefully.

Now, the end of the red train reaches Joe after l_red/30 s, and the end of the blue train reaches him after l_blue/20 s; so l_red/30 = l_blue/20 + 10.

@joe-shmo said
Sorry, incorrect ( but close ).

Examine this derivation more carefully.

Now, the end of the red train reaches Joe after l_red/30 s, and the end of the blue train reaches him after l_blue/20 s; so l_red/30 = l_blue/20 + 10.

Oh! Got them the wrong way 'round, didn't I? l_red/30 + 10 = l_blue/20.

Then you get 20 l_red + 200 = 30 l_blue. And 30 l_blue = 30000 - 30 l_red.

So 20 l_red + 200 = 30000 - 30 l_red <=> 50 l_red = 29800 <=> l_red = 596 and l_blue cannot be found.

@shallow-blue said
Oh! Got them the wrong way 'round, didn't I? l_red/30 + 10 = l_blue/20.

Then you get 20 l_red + 200 = 30 l_blue. And 30 l_blue = 30000 - 30 l_red.

So 20 l_red + 200 = 30000 - 30 l_red <=> 50 l_red = 29800 <=> l_red = 596 and l_blue cannot be found.

Yes you did have them the other way around, but now you have done something else ( check your arithmetic ).

Red is not 596 m, and blue can definitely be found!

@joe-shmo said
Yes you did have them the other way around, but now you have done something else ( check your arithmetic ).

Red is not 596 m, and blue can definitely be found!

Ah. 20 l_red + 6000 = 30 l_blue. So l_red = (30000 - 6000) / 50 = 480, and l_blue = 520. Those numbers make more sense, too.

If I'd done this neatly on paper rather than on the fly in a comment window, I wouldn't have made those mistakes. I hope...

@shallow-blue said
Ah. 20 l_red + 6000 = 30 l_blue. So l_red = (30000 - 6000) / 50 = 480, and l_blue = 520. Those numbers make more sense, too.

If I'd done this neatly on paper rather than on the fly in a comment window, I wouldn't have made those mistakes. I hope...

Correct!