# Question #eaa75

Nov 12, 2017

The question is missing, but ...
The fly will be able to turn around at least six times before the trains crash to annihilate it.

#### Explanation:

Two trains are on the same line, $60$ miles apart.
They are heading towards each other, each traveling at $30 m h$.
Their combined speed is then $30 m p h + 30 m p h = 60 m p h$
This means the trains are $1 h$ apart.

A fly that can travel at $60 m p h$ leaves one engine flying towards the other. So it will take the fly $1 h$ to get from one engine to the other, but only if the two engines are stationary.
Here, the fly is travelling toward the other engine at a rate of $60 m p h + 30 m p h = 90 m p h$, or $1.5 \left(60 m p h\right)$.

So, if we investigate the travel time at the half-way point $= \frac{1}{2} h :$
Each train has moved toward each other by $\frac{1}{2} \left(60 m\right) = 30 m$

The fly has moved toward the other train by $\left(90 m p h\right) \left(\frac{1}{2} h\right) = 45 m$, and will turn back to the original train.

And, if we investigate the travel time at the half-way point here $= \frac{1}{2} \ast \frac{1}{2} = \frac{1}{4} h :$
Each train has moved toward each other by $\frac{3}{4} \left(60 m\right) = 45 m$

The fly has moved toward the other train by $\left(90 m p h\right) \left(\frac{1}{4} h\right) = 22.5 m$, and will turn back to the other train.

And, if we investigate the travel time at the half-way point here $= \frac{1}{2} \ast \frac{1}{2} \ast \frac{1}{2} = \frac{1}{8} h :$

Each train has moved toward each other by $\frac{7}{8} \left(60 m\right) = 52.5 m$

The fly has moved toward the other train by $\left(90 m p h\right) \left(\frac{1}{8} h\right) = 11.3 m$, and will turn back to the original train.

That means the fly will be able to turn around at least six times before the trains crash to annihilate it, depending on where the fly is located vertically amongst protruding surfaces on the trains.