+1521 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?
+34 Will rows for 45 minutes until Grace starts. In that time Will advances \(45 \cdot 50 = 2250\) meters. Every minute they get \(80\) meters closer to each other. At the time Grace starts, they will be \(550\) meters apart. It will take \(\frac{550}{80}\)minutes, which is around \(7\) minutes. They will meet at around \(\boxed{2:52} .\)
+1938 First, let's calculate the distance Will travels before grace even leaves.
Since he traveled for 45 minutes, we have
\((45 min)(50 m/min) = 2250 m\)
So when Grace starts, she and Will have
\((2800 m) – (2250 m) = 550 m\) to cover.
Let's let t be the amount of time it takes for the two to catch up to each other after Grace starts rowing.
We have the equation
\((50)(t) + (30)(t) = 550 \\ 80t = 550 \\ t = 550/80 = 6.875\)
This is approximately 6 minutes and 52 seconds.
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
Thus, our final answer is \(2:51:52\)
Thanks! :)
