Two cyclists started to ride at 8am: 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. They met for the first time at 11 am and each of them turned exactly once before they met for the second time. Find the time of their second meeting.
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2007-08-26 22:24:24 · 4 answers · asked by Ginga-Ninja 1 in Science & Mathematics ➔ Mathematics
When they met the first time, the total distance travelled by both of them together was the distance from A to B. If they each turn once before they met again, they have each travelled the whole distance from A to B plus a bit more, with the two extra bits summing up to be the distance from A to B. So the total distance travelled by both of them is now three times the distance from A to B. This will have taken three times as long as the time to the first meeting, which was 3 hours after they started, so they meet for the second time 9 hours after they started, at 5 pm.
Of course, it's slightly unreasonable to expect any cyclist to maintain a constant velocity for 9 hours.
2007-08-26 22:32:49 · answer #1 · answered by Scarlet Manuka 7 · 2⤊ 0⤋
Calculation:
= 11 + 2(11 - 8)
= 11 + 2(3)
= 11 + 6
= 17
= 17 - 12
= 5 P.M.
2007-08-27 05:41:19 · answer #2 · answered by Jun Agruda 7 · 3⤊ 0⤋
i think at 5 pm
2007-08-27 05:37:04 · answer #3 · answered by Anonymous · 0⤊ 0⤋
2 p.m
2007-08-27 07:38:21 · answer #4 · answered by poonam 1 · 0⤊ 1⤋