+2653 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?
+1926 First, let's calculate how far Will got before Grace even started to row,
We have \(d=st\), where d is distance, s is speed, and t is time.
We have \((45 min)(50 m/min) = 2250 m \). Will rowed 2250m before Grace began.
Now, Will and Grace have to cover \((2800 m) – (2250 m) = 550 m \) together.
We have
\( (50)(t) + (30)(t) = 550 \\ 80t = 550 \\ t = 550/80 = 6.875 \)
6.875 minutes is about 6 minutes and 52 seconds.
The two started at 6:45, so they arrive at about \(6:51:52\) P.M. when they meet.
Thanks! :)
