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

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> 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      Log On


   

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(20055) About Me  (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 1%2F2
an hour at 4mph, and so has a head start of 2 miles, using the formula

DISTANCE = RATE x TIME = %284%29%281%2F2%29=2 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 TIME=%28DISTANCE+BETWEEN+THEM%29%2F%28APPROACH+RATE%29

TIME+=+2%2F1.5+=+4%2F3+=+1%261%2F3 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) About Me  (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