Question

A train approaches a 800 m long tunnel AB.Inside the tunnel there is a rabbit when the train whistles,the rabbit runs.If the rabbit moves to the entrance of tunnel A,the train catches the rabbit exactly at the entrance.If the rabbit moves to exit B,then the train catches the rabbit at the exit.If the speed of the train is 4 times the speed of the rabbit,what is the distance of the point the train whistles from the entrance of tunnel.

Answer 1

Here's my way of solving it.
x__________A______y______B
x is the point where the train is when it whistles, and y is the point where the rabbit is when the train whistles. Let's consider the distance between x and A as D1 and the distance between x and y as D2. The distance between A and B is 800m and the speed of the train, Vt, is 4 times bigger than the speed of the rabbit (Vb), so Vt = 4*Vb.

The first case is that in which the rabbit runs towards A.
The rabbit's equation: D1 - D2 = Vb * T1 (where T1 is the time in which it reaches A)
The train's equation: D2 = 4 * Vb * T1.
Now we multiply the first equation by 4, so that we'll have the same amount on the right side of the "= "sign as in the second. We get:
4 * (D1 - D2) = D2. in the end, 5 * D2 = 4 * D1, so D1 = (5/4) * D2

In the second case, the rabbit runs towards B. So:
AB - ( D1 - D2 ) = Vb * T2
D2 + AB = 4 * Vb * T2.
We multiply the first one by 4 and get:
4 * AB - $ * ( D1 - D2 ) = D2 + AB
3 * AB = D2 + 4 * D1 - 4 * D2.
3 * AB = 4 * D1 - 3 * D2.
We substitute D1 for 5/4 * D2.
3 * AB = (5/4) * 4 D2 - 3 * D2.
so 3 * AB = 2 * D2.
D2 = (3/2) * AB = (3/2) * 800 = 1200m

Answer 2

For the first case rabbit will travel D2 - D1 distance ??

Answer 3

How it is D1 - D2???

Answer 4

Epilepticon are you there?

Answer 5

How T1 will be same when train reaches point A and rabbit reaches point A.

Answer 6

According to Epilepticon assumption,
I derive it as :-
Trains equation of reahing at A :-
D1 = 4 * Vb * T1
Rabbit equation of reaching at point A,
D2 - D1 = Vb * T2
Where T1 = time taken by train to reach A
and T2 = time taken by rabbit to reach A

Answer 7

Am I correct ? Please help.

Answer 8

r t = d
Let x equal the distance the rabbit runs towards entrance A. The train's speed is four times that of the rabbit. Therefore, the train must be at a starting distance of 4x from entrance A.

If the second case the rabbit runs a distance of 800-x to exit B. The train moves a distance of 4x + 800 to intercept the rabbit at exit B. Again, the train moves 4 times any distance that the rabbit runs. The ratio of the train's and the rabbit's distance to B is 4 to 1 respectively.

Solve the following equation for x:\[\frac{4x+800}{800-x}=\frac{4}{1}\]\[x=300 m , 4x=1200 m , \text{The } \text{train } \text{whistles } 1200m \text{ from} \text{ entrance } A. \]

Answer 9

According to Epilepticon assumption, I derive it as :- Trains equation of reahing at A :- D1 = 4 * Vb * T1 Rabbit equation of reaching at point A, D2 - D1 = Vb * T2 Where T1 = time taken by train to reach A and T2 = time taken by rabbit to reach

Answer 10

Why T1 =T2

Answer 11

????

Answer 12

One question for you Epilepticon .If the ratio of speed is 4:1 then time ratio will be 1:4.Correct?
But why you have considered in first part T1 as time for rabbit and train both?

Answer 13

@ mkuma36
The trains's whistle starts a clock, a stop watch so to speak, and the point of each train/rabbit interception stops the clock. The elapsed time of the problem events as measured by any clock, operated by any impartial observer, is identical for both train and rabbit and is not provided in the problem statement, except as a ratio.

The upshot of it all is that the problem solution will be the same whether the speed of the rabbit is that of a snail or that of a Kentucky Derby race horse.