document.write( "Question 154100: a person on a bike leaves an hour before a person on a motorcycle. the person on the bike is travelling at 10 mph. the person on the motorcycle is travelling at 40 mph. how long before the person on the motorcycle overtakes the person on the bike? \n" ); document.write( "
Algebra.Com's Answer #113421 by nerdybill(7384)\"\" \"About 
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a person on a bike leaves an hour before a person on a motorcycle. the person on the bike is travelling at 10 mph. the person on the motorcycle is travelling at 40 mph. how long before the person on the motorcycle overtakes the person on the bike?
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\n" ); document.write( "Let t = time it takes person on motorcycle to overtake
\n" ); document.write( "t+1 = time person on bike traveled
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\n" ); document.write( "Applying the \"distance formula\"
\n" ); document.write( "d = rt
\n" ); document.write( "where
\n" ); document.write( "d is distance
\n" ); document.write( "r is rate or speed
\n" ); document.write( "t is time
\n" ); document.write( ".
\n" ); document.write( "For motorcycle person to overtake bicycle person, the distance traveled by both must be the same:
\n" ); document.write( "40t = 10(t+1)
\n" ); document.write( "40t = 10t + 10
\n" ); document.write( "30t = 10
\n" ); document.write( "t = 10/30
\n" ); document.write( "t = 1/3 hour
\n" ); document.write( "Or, in terms of minutes:
\n" ); document.write( "t = 1/3 * 60 = 20 minutes\r
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