People A,B,C,D, each is at each edge of a square, the square side is 100 meter. A moves to B, B moves to C, C moves to D, D moves to A at speed 1m/s. Simultaneously a bird flies from A to B, then to C, then to D, then to A and so on. The bird meets D for the first time at the 30 second, when will it meet D for the third time?
110 seconds.
ReplyDeleteAftwr meeting D at 30 seconds, the bird has flown 330 m for a speed of 11 m/s.
To catch D, the bird must fly 400 m + the distance travelled by D, in the same time.
Thus,
Tbird = (400 +distD)/11mps
TD = distD/1mps
Solving TD = Tbird gives
distD = 40m - time = 40 sec
Thus, they meet for a second time at 30 + 40 = 70 sec.
Similarly they meet again after
70+ 40 sec = 110 sec.
D has travelled 110 m, the bird has travelled 1220 m.
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ReplyDelete110 seconds
ReplyDelete1st crossing at :- 30 sec
ReplyDelete2nd crossing at :- 40 sec
3rd crossing at :- 40 sec
i.e. a bird will take 110sec for the third time.