The fly flying between two trains

Question

I know this question has been posted many times, but I don't understand it. Two trains travel on the same track towards each other, each going at a speed of 40 kph. They start out 180km apart. A fly starts at the front of one train and flies at 100 kph to the front of the other; when it gets there, it turns around and flies back towards the first. It continues flying back and forth till the two trains meet and it gets squashed.

How far did the fly travel before it got squashed? I must do it John von Neumann's way but I don't get it.. Help maybe ?

Answer 1

The gap between the trains narrows by 80 km each hour, so it'll take 2 hours and 15 minutes before they collide. During that time the fly will travel a distance of $$ 2.25\ hrs \times 100\ km/hr= 225\ km. $$

Answer 2

Here's my explanation:

1. The Two trains are traveling at the same speed of 40 Kilometers/hour
2. Which means they traveled the same distance 180/2 = 90 Kilometers when they met

3. The time to travel 90 Kilometers should also be the same for both the trains:

   40 Km in 1 Hour implies

   90 Km will take (90/40) = 2.25 Hours

4. As the fly was traveling only till the train met, its travel time is 2.25 Hours

5. The flying speed of the fly:

   in 1 Hour it traveled 100 Kms that implies

   in 2.25 Hours it would have traveled 100 * 2.25 = 225 Km

Therefore, the answer to your question is 225 Km

Answer 3

To solve this the Neumann's way, we calculate the time the fly takes in each of its Trips between $A$ and $B$.

Trip $1$ From $A$ to $B$:The fly travels at $100$ kph from $A$ towards the train coming from $B$ at $40$ kph. That means, it is traveling at a relative speed of $140$ kph and it has to cover a distance of $180$ km to meet the train coming from $B$. Time it takes for this Trip $1$ is $(9/7)$ hr. Let us call this $T_1$.

Trip $2$ From $B$ to $A$: In this Trip, the fly does not cover the complete distance of $180$ km, since each train would have covered $(40)*(9/7)$ km by the time the fly completes its Trip $1$. So, the distance the fly has to cover in Trip $2$ is $(180) -(40)*(9/7) -(40)*(9/7)$, which is equal to $(180)*(3/7)$ km. In this Trip, the fly is again traveling at a relative speed of $140$ kph towards the train coming from $A$. Time it takes for this Trip $2$ is $(180)*(3/7)*(1/140)$, that is, $(9/7)*(3/7)$ hr. Let us call this $T_2$.

Continuing the same way, we get

$T_3$ for Trip $3$ From $A$ to $B$ = $(9/7)(3/7)^2$

$T_4$ for Trip $4$ From $B$ to $A$ = $(9/7)(3/7)^3$

And so on. (Remember, the distance the fly covers keeps decreasing from Trip to Trip.)

The total time the fly takes between the trains is:

$(9/7)+(9/7)*(3/7)+(9/7)*(3/7)^2+(9/7)*(3/7)^3+(9/7)*(3/7)^4.......$

The sum to infinity of this Geometric Series is equal to $9/4$ or $2.25$ hrs.

At $100$ kph, the fly therefore travels a total distance of $225$ km.