document.write( "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? \n" ); document.write( "

Algebra.Com's Answer #151687 by Edwin McCravy(20056) $\"\"$ $\"About$

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?\r
\n" ); document.write( "\n" ); document.write( "

\r\n" );
document.write( "When the hornet leaves, the bee has already been flying \r\n" );
document.write( "an hour at 4mph, and so has a head start of 2 miles, using the formula\r\n" );
document.write( "\r\n" );
document.write( "DISTANCE = RATE x TIME =  miles head start.\r\n" );
document.write( "\r\n" );
document.write( "The hornet's approach rate is the difference in their speeds, \r\n" );
document.write( "\r\n" );
document.write( "5.5mph - 4mph = 1.5mph\r\n" );
document.write( "\r\n" );
document.write( "(That is, the hornet gains a mile and a half on the bee every hour.)\r\n" );
document.write( "\r\n" );
document.write( "Now using \r\n" );
document.write( "\r\n" );
document.write( " hours\r\n" );
document.write( "\r\n" );
document.write( "or 1 hour 20 minutes.  The hornet left at 12:30PM and closed up\r\n" );
document.write( "the 2-mile gap between him and the bee 1 hour and 20 minutes \r\n" );
document.write( "later, so that would have been at 1:50PM\r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( " \n" ); document.write( "

\n" );