document.write( "Question 562282: at 7:00 am joe starts jogging at 6mi/h . At 7:10 Ken starts off after him. How fast must ken run in order to overtake him at 7:30 \n" ); document.write( "

Algebra.Com's Answer #364526 by mananth(16946) $\"\"$ $\"About$

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Joe speed = 6 mph
\n" ); document.write( "Ken speed = x mph\r
\n" ); document.write( "\n" ); document.write( "Joe starts 10 minutes early
\n" ); document.write( "so he has run 6*10/60 = 1 mile before ken starts\r
\n" ); document.write( "\n" ); document.write( "distance Ken has to run to catch up = 6-1=5 miles\r
\n" ); document.write( "\n" ); document.write( "catchup time = 20 minutes ( from 7:10 to 7:30)= 2/3 hours\r
\n" ); document.write( "\n" ); document.write( "speed = d/t\r
\n" ); document.write( "\n" ); document.write( "speed = 5/(2/3)\r
\n" ); document.write( "\n" ); document.write( "speed = 5 *3/2\r
\n" ); document.write( "\n" ); document.write( "speed = 15/2 = 7.5 mph\r
\n" ); document.write( "\n" ); document.write( "CHECK
\n" ); document.write( "catchup time = 2/3
\n" ); document.write( "Joe started 1/6 hour early
\n" ); document.write( "so he ran for 2/3 +1/6 = 5/6 hours
\n" ); document.write( "his speed = 6 mph
\n" ); document.write( "5/6 hour *6 = 5 miles\r
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