document.write( "Question 63001: 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
\n" ); document.write( "will John catch up with Martina? \n" ); document.write( "
Algebra.Com's Answer #43917 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
\n" ); document.write( " later, John leaves, driving at the rate of 48 mi/h. At what time will John
\n" ); document.write( " catch up with Martina?
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\n" ); document.write( "One thing to remember about these \"catch-up\" problems is when it occurs, both
\n" ); document.write( "parties will have traveled the same distance. Make a distance equation>
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\n" ); document.write( "Let t = M's time in hrs from 9 AM
\n" ); document.write( "Then (t-2) = J's time when he overtakes M
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\n" ); document.write( "Dist = speed*time
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\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 would be 1 pm when J catches up
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\n" ); document.write( "Check our solution using the dist equation:
\n" ); document.write( "48(2) = 24(4)
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\n" ); document.write( "Did this make sense to you??\r
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