Question 1207791
<pre>
The best way to really understand and explain this problem is to have the bees
flying in a circle of circumference twice 150 or 300, so that each semi-circle
from A to B is 150 m.  We are not told the bees must fly in a straight line, so
we are certainly free to assume they fly in this circle. It is 150 circular
meters from A to B. (around the circumference of each semi-circular path.

The slower bee flies clockwise from A, and the faster bee flies
counter-clockwise from B.  

In the circles below, the two bees are the dots at the end of the arcs through
which the bees have just traveled.  The slower bee is green, and the faster one
red.

In the first circle and third circles below, the bees are meeting. In the
middle circle they are 150 circular meters apart. (You can think of them
as 180<sup>o</sup> apart, if you like.)  


{{{drawing(200,200,-1.2,1.2,-1.2,1.2,

red(arc(0,0,1.9,-1.9,0,113)), green(arc(0,0,1.8,-1.8,113,180)),
locate(-1.1,.1,A),locate(1.06,.1,B),


green(circle(-.38,.8,.03)), red(circle(-.385,.86,.03)),


circle(0,0,1), circle(-.3826834324,.9238795325,.01) )}}}{{{drawing(200,200,-1.2,1.2,-1.2,1.2,

red(arc(0,0,1.9,-1.9,113,226)), green(arc(0,0,1.8,-1.8,45,113)),

locate(-1.1,.1,A),locate(1.06,.1,B),

green(circle(.65,.61,.03)), red(circle(-.6,-.7,.03)),


circle(0,0,1), circle(-.3826834324,.9238795325,.01) )}}} {{{drawing(200,200,-1.2,1.2,-1.2,1.2,

red(arc(0,0,1.9,-1.9,226,339)), green(arc(0,0,1.8,-1.8,344,405)),
locate(-1.1,.1,A),locate(1.06,.1,B),



green(circle(.86,-.26,.03)), red(circle(.91,-.3,.03)),



circle(0,0,1), circle(-.3826834324,.9238795325,.01) )}}}

Ikleyn above did not understand why I said this: 

The amount of time from when they are 150 m apart until the next time they will
be 0 m apart

----will always be the same as---- 

The amount of time from when they are 0 m apart until the next time they will
be 150 m apart

She wasn't thinking of the bees as if they were flying in a circular or
curved path as I was.   

But as you see, the bees travel the same distance in all three circles, so
it will take the same length of time.  

In the first circle, their rate of approach is 3+5=8 m/s. So their 150 m
distance of separation will shrink to 0 in 150/8 = 18.75 seconds.

So it will take another 18.75 seconds until they are 150 m apart. 

Then it will take another 18.75 seconds for them to get 0 m apart again. 

Answer: 18.75 + 18.75 = 37.5 more seconds.

Edwin</pre>