document.write( "Question 59388This question is from textbook
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document.write( ": Science and medicine. Martina leaves home at 9 A.M., bicycling at a rate of
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document.write( "24 mi/h. Two hours later, John leaves, driving at the rate of 48 mi/h. At what time
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document.write( "will John catch up with Martina? \n" );
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Algebra.Com's Answer #40718 by funmath(2933)![]() ![]() ![]() You can put this solution on YOUR website! Science and medicine. 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? \n" ); document.write( ": \n" ); document.write( "The distance formula is: \n" ); document.write( ": \n" ); document.write( "Martina's r=24 mi/h \n" ); document.write( "Let Martina's time be: t (we don't know what it is) \n" ); document.write( "Then Martina's distance d=24t \n" ); document.write( ": \n" ); document.write( "John's r=48 mi/h \n" ); document.write( "John left two hours later so he traveled 2 hours less than Martina: t-2 \n" ); document.write( "Then John's distance d=48(t-2) \n" ); document.write( ": \n" ); document.write( "John Catches up to Martina, so their distances are equal. \n" ); document.write( "Problem to solve: \n" ); document.write( "24t=48(t-2) \n" ); document.write( "24t=48t-96 \n" ); document.write( "-48t+24t=-48t+48t-96 \n" ); document.write( "-24t=-96 \n" ); document.write( "-24t/-24=-96/-24 \n" ); document.write( "t=4 \n" ); document.write( "That means that Martina traveled for 4 hours. \n" ); document.write( "She left at 9am, four hours later makes it 1pm. \n" ); document.write( "Happy Calculating!!! \n" ); document.write( " |