# 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 ->  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

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 Click here to see ALL problems on Linear Equations And Systems Word Problems 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(8895)   (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(12655)   (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