David is swimming upstream from a dock at a constant rate. After he has swum 400 metres, he meets Julie who is floating downstream on an air-mattress. He contiunes to swim upstream for a further 10 minutes before turning round and swimmind back to the dock, which he reaches ar the same time as Julie. Assume that David's swimming is constant relative to the water and that the current has constant speed of x metres per minute. Find x.
2007-09-03 01:14:00 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Assume that David's swimming speed in still water is y meters per minute. That is also his swimming speed relative to water. So David had swum a total of 400 + 10(y-x) meters upstream. It then took him {400 + 10(y-x)}/(y+x) min to swim back. Hence after David met with Julie, It took Julie 400/x min to reach the dock, while took David 10 + {400 + 10(y-x)}/(y+x) to reach the dock. But these two time periods must be the same. Therefore:
400/x =10 + {400 + 10(y-x)}/(y+x)
Timing both sides by x(x+y) and combine identical terms we get:
400y = 20xy, or
x = 20
2007-09-03 09:05:26 · answer #1 · answered by Hahaha 7 · 0⤊ 0⤋