Homework Help, trains

A physics teacher, whose name will not be mentioned to protect the guilty, came to a narrow railroad bridge and began to cross it. He had crossed 3/8 of the bridge length when a whistle warned him of an approaching train. Being a bright individual, he quickly evaluated his alternatives. If he ran back at 10 m/s he would exit the bridge just before the train arrived. If he ran forward at 10 m/s he would exit the bridge just before the train caught him. How fast was the train going?

I have an answer but I was wondering if I was off or not. I'm also wondering how someone else would go about solving this.

Thank you for your time.

This one is a bit strange - intuition suggests you will also need to know the length of the bridge but I may be wrong.

As far as the available information goes : to go back the way he has come will require that he travel 3/8ths bridge length again whereas if he keeps going then he will have to travel 5/8ths bridge length.
If he continues, when he is 6/8ths of the way over the train will be at the start of the bridge. Thus the train takes the same length of time to cross the entire bridge as the boffin takes to cover 2/8ths of it.
Wow - it is going at 40m/s.
Hope this is right.

if the length of the Bridge is I, and the teacher is at 3/8I, and Train is X metes away from him when it releases a noise, and the noise (say the whistle) travels at a constant speed of V m/s , and if "t" is the time it takes for the train whistle to reach the teacher, then the train approaches the teacher by V * t meters during the time it takes the noise to travel the Vt meters. So lets make some obvious assumptions:

Since the teacher can outrun the train by either going inside the cave or out , it is only logical that the train releases the whistle when it is at a considerable distance on the other side of the bridge. If it was , per se, to release its noise once inside the bridge then the teacher would be dead because if he moved by ten meters per second toward the train, the train would just cut him into fine pieces. Assuming that earth is a constant frame of refrences, the train is moving at V m/s toward the teacher who WE ASSUME to be motionless until he hears the noise. Lets draw a hypothetical diagram:

Ok, at the very moment the train releases a noise, the following is the hypothetical setting:
...............................................................<-----------5/8I--------><-3/8I--->
))))))))))))))~~~~~~~~~~~~~~~~~~~|------------------------=P----------|
.................<-------------------Y------------->
.................<-------------------------------X----------------------------->
and lets say the it takes "t" seconds for the train whistle to reach the teacher. During this time the train moves a distance of Vt. Lets call the original distance of the train from the tunnel Y . After t seconds the train is now Y - Vt meters away from the brige and if we call X the original distance of the train from the teacher they are now X - Vt meters away.

At the very moment the teacher hears the noise of the train whistle the following is a representive of the distances:
........................................<------------5/8I----------><-3/8I--->
))))))))))))~~~~~~~~~~~|---------------------------=P--------|
...............<---- (Y - Vt)---->
...............<---------------------------(X - Vt)------------->


Now, assuming that the teacher is a super-genius who can calculate and find answers in an instance (meaning wastes no time to make a decision) he then realises as you stated that he should move at 10 m/s. Now lets make assumptions. If he is to scape the train the the train should be barely touching him when he reaches the end of the bridge or should barely miss him if he moves back outside of the bridge. Lets say while the teacher runs like hell to avoid death it takes him exactly T seconds to avoid death. Assuming the train driver is asleep , dead or just blind, and doesnt stop the train is still moving at V m/s and would go a distance of VT seconds till the very moment that the teacher escapes. Whether the teachers goes toward the the train or run away from it it would be the same distance for the train. Now lets divide our assumptions into two groups:
_______________________
Condition #1)

The teacher decides to run away by returning the 3/8 of the bridge length. During this time the train moves a distance of VT and the teacher moves at a constant velocity of Z m/s (i dont use 10m/s yet because I am not sure it is right) so during this very same time he woudl make about ZT meters run. Since the distance of the teacher from the begginning of the tunnel is 3/8I where I is the length of the tunnel we can say that he goes ZT meters or 3/8I or 3/8 I = ZT meters. Assuming that by Z he barely misses the train then during this very same time the train has come a distance of : Y -Vt + I. Since both events take place simultaneously then the train's velocity must be Y - Vt + I / T , where T is the time it takes the teacher to escape the train.

Now we have two relationships:

(1) ZT = 3/8 I , (2) (Y - Vt + I ) / T = V

using both (1) and (2) we have:

(Y - Vt + I) * 8 Z / 3 I = (8YZ - 8ZVt + 8IZ) / 3I = V

_______________________
Condition #2

This time the teacher runs toward the train. The train should barely miss it so the train must move only Y - Vt meters and the teacher has to move a distance of 5/8I at the very same speed of Z meters per second. The train moves at a constant velocity of V meters per second and it goes a distance of Y - Vt in T' so

V = Y - Vt / T'

and also teacher goes 5/8I in T' seconds so Z = 5I/8T' or T' =5I/8Z

Having T' in common we can then device the following formula:

V = (Y - Vt) 8Z / 5I = 8ZY - 8ZVt / 5I

Since in both conditions the train moves at a constant speed and so does the teacher we have:

V= (8ZY - 8ZVt) / 5I = (8YZ - 8ZVt + 8IZ) / 3I

Lets now simplify :

(8ZY - 8ZVt ) / 5 = (8YZ - 8ZVt + 8IZ) / 3


1.6ZY - 1.6ZVt = 2.7ZY - 2.7ZVt + 2.7IZ


1.6ZY - 1.6ZVt - 2.7ZY + 2.7ZVt - 2.7IZ = 0

-1.1ZY + 1.1 ZVt - 2.7IZ = 0

Z ( 1.1Vt - 1.1Y - 2.7I ) = 0

Z = 0 m/s !!!

As suspected I think it is physically impossible for the teacher to move to the both sides of the tunnel or the bridge and still Barely escape the train. He might esacpe the train Barely by running toward it and miss it with time to spare by running back which would make sense but this is not the case of the questoin or so it is my understanding of it.

No, the speed of sound will not matter.

I will do it with the speed of sound included to show it is still 40 m/s

let a= speed of sound in m/s
let y= the train's distance from the bridge when the whistle is sounded
let z= the train's distance from the bridge when the teacher hears the whistle
let r= the rate at which the train is traveling
let x= the length of the bridge

now the dist that the sound must travel to reach the teacher = y+(3*x/8)
thus the time the sound must travel = (y+(3*x/8))/a
in this amount of time the train travels a dist of rate*time or
(r/a)*(y+(3*x/8)) this is = y-z
so now the dist that the train is from the tunnel = z = y- (r/a)*(y+(3*x/8))

now we introduce our 2 conditions. Essentially they are just saying that the times required for the teacher to move 3*x/8 at 10 m/s and the train to move z at r are equal. Also that the time for the teacher to move 5*x/8 at 10 m/s and the train to move z+x are equal. So we now have 2 eqns with each being

(dist/rate) of teacher = (dist/rate) of train

putting in variables

((3*x)/(8*10))=z/r= (y- (r/a)*(y+(3*x/8)))/r

and

((5*x)/(8*10))=(z+x)/r=(x+y- (r/a)*(y+(3*x/8)))/r

distributing gives us this

(3x/80)=y/r-y/a+3x/8a

and

5/x/80=x/r+y/r-y/a+3x/8a

take the 2nd eqn minus the first and you have

2x/80=x/r

thus r=40=rate of train

i tried to draw it but it kept moving the lines on me


sorry

ya, It was already said to be 40 but he wants the conditions where either way if he runs left or right he should barely miss the train...or is that a misunderstanding? But on the other hand, I most likely made a mistake but well, two ppl say 40 it is the right answr then.

no...thats the correct understanding and thats how i solved. Logically it does make sense.

If he runs at the train to escape the bridge he has to run a shorter distance to safety but his time is shortened because the train and he are moving towards each other. If he runs away from the train, the teacher must go a longer distance but he buys himself more time by running away from the train.