document.write( "Question 128131: Two cyclists start biking from a trail’s start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking? \n" ); document.write( "

Algebra.Com's Answer #93829 by ptaylor(2198) $\"\"$ $\"About$

You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "Let t=time that will pass before the second cyclist catches up with the first from the time the second cyclist started biking\r
\n" ); document.write( "\n" ); document.write( "Now we know that when they have both travelled the same distance, the second cyclist will have overtaken the first\r
\n" ); document.write( "\n" ); document.write( "Distance first cyclist travels=6t+6*3=6t+18 (note: the first cyclist has travelled 6*3 or 18 miles before the second cyclists starts)\r
\n" ); document.write( "\n" ); document.write( "Distance second cyclist travels=10t\r
\n" ); document.write( "\n" ); document.write( "Now we know that these two distances must be equal, so:\r
\n" ); document.write( "\n" ); document.write( "6t+18=10t subtract 6t from both sides\r
\n" ); document.write( "\n" ); document.write( "6t-6t+18=10t-6t collect like terms\r
\n" ); document.write( "\n" ); document.write( "18=4t divide both sides by 4\r
\n" ); document.write( "\n" ); document.write( "t=4.5 hrs\r
\n" ); document.write( "\n" ); document.write( "Ck
\n" ); document.write( "6*4.5+18=10*4.5
\n" ); document.write( "27+18=45
\n" ); document.write( "45=45\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor\r
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