Two cyclists start biking from a trail 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 beforenthe second cyclist catches up with the first from the time the second cyclist started biking?

The first cyclist has a 3 hour head start--and in those three hours, has traveled 3 X 6 = 18 miles. The question now is: how long will it take for the second cyclist to travel 18 miles more than the first cyclist. The second cyclist will gain 10 - 6 = 4 miles / hour so 18 / 4 = 4.5 hours the second cyclist will catch up in 4.5 hours after starting. with equation goodness: distance = speed  X time (where t is time after the second cyclist starts) 18 = 10t - 6t = 4t 18/4 = t = 4.5