document.write( "Question 252492: Two cyclist start biking from a trail's start 3 hours apart, The second cyclist travels @ 10mph and starts 3 hourss after the first cyclist who is traveling @ 6mph. How much time will pass before the second cyclist catches up with the first from the time the second cyclist srtarted biking? \n" ); document.write( "
Algebra.Com's Answer #184457 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! Two cyclist start biking from a trail's start 3 hours apart, \n" ); document.write( " The second cyclist travels @ 10mph and starts 3 hours after the first cyclist who is traveling @ 6mph. \n" ); document.write( " 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( ": \n" ); document.write( "Let t = travel time of the 2nd cyclist \n" ); document.write( "then \n" ); document.write( "(t+3) = travel time of the 1st cyclist \n" ); document.write( ": \n" ); document.write( "When the 2nd catches the 1st, they will have traveled the same distance \n" ); document.write( ": \n" ); document.write( "Write a distance equation: dist = speed * time \n" ); document.write( ": \n" ); document.write( "2nd cyclist dist = 1st cyclist dist \n" ); document.write( "10t = 6(t+3) \n" ); document.write( "10t = 6t + 18 \n" ); document.write( "10t - 6t = 18 \n" ); document.write( "4t = 18 \n" ); document.write( "t = \n" ); document.write( "t = 4.5 hrs for the 2nd cyclist to catch the 1st \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution (1st cyclist travel time = 7.5 hr) \n" ); document.write( "10 * 4.5 = 45 mi \n" ); document.write( "6 * 7.5 = 45 mi \n" ); document.write( " |