+1678 Will and Grace are canoeing on a lake. Will rows at $50$ meters per minute and Grace rows at $30$ meters per minute. Will starts rowing at $2$ p.m. from the west end of the lake, and Grace starts rowing from the east end of the lake at $2{:}45$ p.m. If they always row directly towards each other, and the lake is $2800$ meters across from the west side of the lake to the east side, at what time will the two meet?
+1277
Will and Grace are canoeing on a lake. Will rows at $50$ meters per minute and Grace rows at $30$ meters per minute. Will starts rowing at $2$ p.m. from the west end of the lake, and Grace starts rowing from the east end of the lake at $2{:}45$ p.m. If they always row directly towards each other, and the lake is $2800$ meters across from the west side of the lake to the east side, at what time will the two meet?
Will has a 45 minute head start, so he's already traveled (45 min)(50 m/min) = 2250 m
before Grace even begins to row.
So, the actual distance of closure for Will and Grace is (2800 m) – (2250 m) = 550 m
rate times time equals distance
Since they meet, they will both row the same amount of minutes. Call it t.
Therefore, the setup is (50)(t) + (30)(t) = 550
80t = 550
t = 550/80 = 6.875
The 6 is minutes and the .875 is the fraction of a minute
(60 sec/min)(0.875 min) = 52.5 sec ... let's round that to 52 sec
So the clock time they meet is 6 min 52 sec after Grace starts.
Grace started at 2:45, so add 6 min 52 sec and the clock will read 2:51:52 pm when they meet
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