document.write( "Question 76998: Martina leaves home at 9 A.M., bicycling at a rate of
\n" ); document.write( "24 mi/h. Two hours later, John leaves, driving at the rate of 48 mi/h. At what time will John catch up with Martina?
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Algebra.Com's Answer #55177 by ankor@dixie-net.com(22740)\"\" \"About 
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Martina leaves home at 9 A.M., bicycling at a rate of 24 mi/h. Two hours later,
\n" ); document.write( " John leaves, driving at the rate of 48 mi/h. At what time will John catch up
\n" ); document.write( " with Martina?
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\n" ); document.write( "We know when J catches up with M, they will have traveled the same distance;
\n" ); document.write( " we can make a simple equation from that fact:
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\n" ); document.write( "let t = the time M on the road when John catches up
\n" ); document.write( "Then (t-2) = time J is on the road
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\n" ); document.write( "Distance = speed * time:J
\n" ); document.write( "J's dist = M's dist
\n" ); document.write( "48(t-2) = 24t
\n" ); document.write( "48t - 96 = 24t
\n" ); document.write( "48t - 24t = + 96
\n" ); document.write( "24t = 96
\n" ); document.write( "t = 96/24
\n" ); document.write( "t = 4 hrs from 9 am is 1 pm
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\n" ); document.write( "Check using dist:
\n" ); document.write( "M: 4*24 = 96
\n" ); document.write( "J: 2*48 = 96
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