[maths challenge] maths query...[Rates of change..??] HELP

An engineer is directed to a faulty signal, one quarter of the way into a tunnel. Whilst there, he is warned of a train heading towards the tunnel entrance. The engineer can run 12 mph and can either run back into the tunnel entrance or forward to the exit. In either case, the engineer and the front of the train would reach the entrance or exit together. What is the speed in mph of the train?

Heres how i've visualised it:

Now for the engineer running to the entrance:
Notation:
X[t]=displacement of train
X[e]=displacement of engineer

We also know:
V[e] = d(X[e])/dt = 12
we need
V[t] = d(X[t])/dt = ?

Displacement of both engineer and train:
X[e]=x/4
X[t]=x+y

4X[e]=x
hence...
X[t]=4X[e]+y

d(X[t])/d(X[e]) = 4

using the chain rule:

d(X[t])/d(X[e]) * d(X[e])/dt = d(X[t])/dt
4 * 12 = 48mph

But the answer is incorrect.....
Where have I gone wrong and how can I go about doing the question, i've noticed I never used the scanerio where the engineer runs toward the exit...

Reply 1

Kolya

What is the answer? Is it 24mph?

Reply 2

The Bachelor

You requested I try solving it:

Spoiler

Reply 3

majikthise

As to why your method didn't work, isn't the train (and hence the "y" distance) meant to be on the left hand side of the diagram?

Reply 4

fusionskd

Whops made a slight error, this is how it should look like:

Notation:
X[t]=displacement of train
X[e]=displacement of engineer

Considering running to the exit:
X[t] = x+y
X[e] = 3x/4

hence...
X[t] = y + 4X[e]/3
d(X[t])/d(X[e]) = 4/3

using the chain rule:

d(X[t])/d(X[e]) * d(X[e])/dt = d(X[t])/dt
4/3 * 12 = 16mph

But the answer is incorrect nonetheless...wheres my error??

Reply 5

Kolya

Ancient Runes solution is far better than mine.

Reply 6

Kolya

fusionskd

hence...
X[t] = y + 4X[e]/3
d(X[t])/d(X[e]) = 4/3

I don't understand this step. You have given no explanation of your Mathematical argument.

Reply 7

The Bachelor

Doing it your way:

Remember that displacement is distance relative to some origin. (Let's say this origin is at the tunnel's entrance and rightwards is positive.) The train starts somewhere left of the tunnel, say it starts distance y left of the tunnel.

Then

X_{t}[t]=-y+V_tt

for the train.

If the engineer goes leftwards: his displacement will be

X_{e}[t]=\frac{1}{4}x-V_et

If the engineer goes rightwards: his displacement will be:

X_{e}[t]=\frac{1}{4}x+V_et

Reply 8

fusionskd

Lusus Naturae

I don't understand this step. You have given no explanation of your Mathematical argument.

X[t] = x+y
X[e] = 3x/4 {making x the subject x=4X[e]/3 }
plugging into the first eqn

X[t] = 4X[e]/3 + y

and differentiating with respect to X[e]

d(X[t])/d(X[e]) = 4/3