# SOLUTION: 2. Two cyclists start biking from a trial and start 3 hours apart. The 2nd cyclist travels at 10 mph and starts 3 hours after the 1st cyclist who is traveling at 6 mph. How much ti

Algebra ->  Algebra  -> Matrices-and-determiminant -> SOLUTION: 2. Two cyclists start biking from a trial and start 3 hours apart. The 2nd cyclist travels at 10 mph and starts 3 hours after the 1st cyclist who is traveling at 6 mph. How much ti      Log On

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 Question 315060: 2. Two cyclists start biking from a trial and start 3 hours apart. The 2nd cyclist travels at 10 mph and starts 3 hours after the 1st cyclist who is traveling at 6 mph. How much time will pass before the 2nd cyclist catches up with the 1st from the time the 2nd cyclist started biking? Please help me with this one i got stuck Thank youAnswer by ankor@dixie-net.com(15622)   (Show Source): You can put this solution on YOUR website!The 2nd cyclist travels at 10 mph and starts 3 hours after the 1st cyclist who is traveling at 6 mph. How much time will pass before the 2nd cyclist catches up with the 1st from the time the 2nd cyclist started biking? : Let t = travel time of the 2nd cyclist then (t+3) = travel time of the 1st cyclist : When the 2nd overtakes the 1st, they will have traveled the same distance ; Write distance equation; dist = speed * time 2nd dist = 1st dist 10t = 6(t+3) 10t = 6t + 18 10t - 6t = 18 4t = 18 t = t = 4.5 hrs for the 2nd cyclist to catch the first ; : Check solution by finding distance of each, should be equal: 10*4.5 = 45 mi 6 *7.5 = 45 mi