SOLUTION: A bee leaves S at noon and flies at 4mph on a straight line towards T. One half hour after the bee leaves S, a hornet leaves Sheading towards T along the same straight line, flyin

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Question 201473: A bee leaves S at noon and flies at 4mph on a straight line towards T. One half hour after the bee leaves S, a hornet leaves Sheading towards T along the same straight line, flying at 5.5mph. At what time will the hornet catch the bee?
Found 2 solutions by Edwin McCravy, rfer:
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
A bee leaves S at noon and flies at 4mph on a straight line towards T. One half hour after the bee leaves S, a hornet leaves Sheading towards T along the same straight line, flying at 5.5mph. At what time will the hornet catch the bee? 

When the hornet leaves, the bee has already been flying 
an hour at 4mph, and so has a head start of 2 miles, using the formula

DISTANCE = RATE x TIME =  miles head start.

The hornet's approach rate is the difference in their speeds, 

5.5mph - 4mph = 1.5mph

(That is, the hornet gains a mile and a half on the bee every hour.)

Now using 

 hours

or 1 hour 20 minutes.  The hornet left at 12:30PM and closed up
the 2-mile gap between him and the bee 1 hour and 20 minutes 
later, so that would have been at 1:50PM

Edwin
Answer by rfer(16322)   (Show Source): You can put this solution on YOUR website!
d=rt
2= lead
1.5 difference in rate
2=1.5t
t=1.33 hr
t=1hr 20min
they will meet at 1:50pm
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