+826 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?
+1894 So first off, let's calculate the distance Will fovers before Grace even begins to row.
We have \((45 min)(50 m/min) = 2250 m \).
When Grace starts to row, the two have to cover \((2800 m) – (2250 m) = 550 m \)
Since they are rowing at eachother at the same time, if we let t be the time it takes, we have \( (50)(t) + (30)(t) = 550 \).
Solving this, we get
\( 80t = 550 \\ t = 550/80 = 6.875 \)
This rounds to approximately 6 minutes and 52 seconds.
This means in addition to the 45 minutes Will takes at the beggining, they will take an additional 6 minutes and 52 seconds to meet.
Now, we some math to find that they meet at \(2:51:52\) P.M.
Thanks! :)
+1894 So first off, let's calculate the distance Will fovers before Grace even begins to row.
We have \((45 min)(50 m/min) = 2250 m \).
When Grace starts to row, the two have to cover \((2800 m) – (2250 m) = 550 m \)
Since they are rowing at eachother at the same time, if we let t be the time it takes, we have \( (50)(t) + (30)(t) = 550 \).
Solving this, we get
\( 80t = 550 \\ t = 550/80 = 6.875 \)
This rounds to approximately 6 minutes and 52 seconds.
This means in addition to the 45 minutes Will takes at the beggining, they will take an additional 6 minutes and 52 seconds to meet.
Now, we some math to find that they meet at \(2:51:52\) P.M.
Thanks! :)
