Date: 08/31/2003 at 00:33:43 From: Sami Subject: time Two cyclists started to ride at 7am: one from A to B, the other from B to A. Each of them rode at a constant speed along the same road and when each arrived at the terminal point, immediately turned back. After they met for the 1st time, each of them turned exactly once before they met for the 2nd time at 6pm. Find the time of their 1st meeting.
Date: 08/31/2003 at 20:36:53 From: Doctor Greenie Subject: Re: time Hi, Sami-- Let C be the point where the two cyclists meet the second time. Let the distance AC be x and the distance BC be y. The two cyclists cycle for 11 hours (7am until 6pm) before meeting the second time. In that time, the cyclist who started from A has traveled from A to C, from C to B, and from B to C, a total distance of (x+2y). In the same time, the cyclist who started from B has traveled from B to C, from C to A, and from A to C, a total distance of (2x+y). You now know the time each cyclist traveled and the distance each one traveled, so you can determine expressions for the rate for each cyclist. You can also determine the total distance (in terms of x and y) the two cyclists needed to travel before meeting the first time. And now you know how far the two cyclists together had to travel before meeting the first time; and you know the rate of each cyclist--so you can determine the amount of time they had to travel before meeting that first time.... I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/
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