A man walks through a railway tunnel. Exactly 1/4 of the way through, he hears a train approach from behind.?

Question

He wonders whether to run back in the direction he came from (which is towards the approaching train), or forwards to the end of the tunnel. What he doesn't know is that in each case, running at a constant speed of 18mph, he will reach the respective mouth of the tunnel at the exact same time as the train does. What is the speed of the train ?

Answer 1

To make this easy, let's assume that the tunnel is 72 miles long (and that the man has INCREDIBLE hearing and endurance).

This means that he is exactly 18 miles into the tunnel and it will take him 1 hour to return to the back entrance of the tunnel. This also means that the train is one hour from the mouth of the tunnel.

Now if he is 18 miles from the entrance of the tunnel, that means he is 54 miles from the further exit ahead of him. It will take him 3 hours to run that distance. After the man has run ahead for 1 hour he is half-way through the tunnel, and the train will have reached the mouth of the tunnel behind him. If they are to arrive at the end of the tunnel at the same time, that means the train will cover twice the distance (72 miles vs 36 miles) in the same time (2 hours). Thus, the train is traveling at...36 mph (18mph x 2).

You actually don't need the distances, but it makes it more concrete. Just think: it takes the same time for the train to get to the entrance of the tunnel as it does for the man to run 1/4 of the distance of the tunnel. So if he runs forward 1/4 of the distance, he will be half-way when the train reaches the entrance...then see above.

Answer 2

Is it 54mph? I'm going to check my working, but that's my first stab. Okay, no, it's 36 mph- all the details may be found below the dotted line....
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I've just 'worked' with my flatmate for about an hour on this, and I'm going to try to reproduce his workings. (He did all the work, and I sat there with my brain on the blink).

We called the exit of the tunnel L, the nearest mouth of the tunnel 'O'. The distance of the train when the man starts running was M. The time at which the train and the man could reach the nearest mouth of the tunnel was marked Ta, and the train and the man would reach the furthest mouth of the tunnel at Tb. The position of the man in the tunnel when he hears the train was L/4. The speed of the train is Vt and the speed of the man is Vm
I'll try to sketch the little chart we came up with;


M____________'O', Ta_____ L/4_________________'L', Tb


We knew that the train will intersect with the position of the man at Ta, expressed thus: -M+ Vt xTa = L/4 - Vm x Ta. = 'O'
' i' 'iii'
For Tb the expression was: -M + Vt x Tb = L/4 + Vm x TB = 'L'
'ii' 'iv'

To establish an expression for the velocity of the train, we subtracted 'i' from 'ii'. Vt x Tb - Vt x Ta = 'L'- 'O'

which came out as Vt = 'L'/ Tb- Ta

Then we reached this stage: As the man is a 1/4 of the way down the tunnel, L/4 = Vm x Ta (to get to 'O')
3/4 L = Vm x Tb (to get to 'L')

Then this took place: Vt = L/ Tb- Ta

whilst Tb= 3'L'/4Vm and Ta= L/4Vm

Therefore: Vt =L/ ( 3L/4Vm) - (L/4VM)

which is: 1/ (3/4Vm0 - (1/4Vm)

which is: 1/ (1/2vm)

Therefore Vt= 2vm - the speed of the train is twice that of the man! Or 36 mph... we tested that with another equasion, one which I will not attempt to reproduce here due to time limitations eg, I've got other stuff to do, and it seemed to be correct.

Answer 3

I'll use the following representations:

time taken for both to reach the entrance of tunnel: t1
time taken for both to reach the exit of the tunnel: t2
distance of the train from the entrance of the tunnel: d
speed of the train: s

t1 = d / s = 0.25 / 18

t2 = (d + 1) / s = 0.75 / 18

notice that:

t2 = 3 x t1

So we can write:

(d + 1) / s = 3d / s

d + 1 = 3d

1 = 2d

so: d = 1/2

Put d = 1/2 back into the very first equation and solve for s:

1 / 2s = 0.25 / 18

18 / 2s = 0.25

18 = 0.5s

So: s = 36

The speed of the train is 36 mph.

Answer 4

It takes the man the same amount of time to travel 3/4 the length of the tunnel as it takes the train to travel 3/2 the length of the tunnel. Similarly, it takes the man the same amount of time to travel 1/4 the length of the tunnel as it takes the train to travel 1/2 the length of the tunnel.

Clearly, the train is going twice as fast as the man, or 36 mph.

Answer 5

I guess you mean that he hears the train as it reaches the tunnel mouth (if we don't assume this there is no possible answer).
The train travels 4/3 times the distance the man travels and must be going 4/3 times faster than the man. 4/3 * 18 = 24 mph

Answer 6

In order to calculate the velocity of the oncoming train moving at a constant speed, we need to know the distances or times involved so we can use the vdt triangle.

If there is another way to work it out, I couldn't give a fcuk as he should not be trespassing on a railway. Serves him right.

Answer 7

If x is distance of train from tunnel and s is lenght of tunnel,t1 is time taken for man to go back where he came from ,t2 time to go to the otheer end of tunnel and u the speed of the train then:
s/4=18*t1 (1), 3s/4=18*t2. so t2=3*t1
Also time train needs to go the beggining of tunnel is
x=u*t1 (2) and to go to the end of the tunnel x+s=u*t2=u*3t1
So: x+s=3x and so s=2x
Replacing in equation (1) x/2=18*t1 i.e. x=36*t1and again replacing this result in equation (2) we get
36*t1=u*t1 i.e. u=36mph
So the answer is 36mph

Answer 8

If this is in UK the guy has got plenty of time to walk to far end of tunnel because i doubt if the train would be moving.

Answer 9

to answer this question i need to know how far the train is from the mouth of the tunnel when this guy first hears it approaching. if i dont have that i cant solve it.

but i like this question... interesting stuff man.

looks like the person before me got it! good stuff man! give him the 10 points.

Answer 10

Is the guy a blond? Gotta be, to be walking through a train tunnel.