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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?

 Jul 21, 2024
 #1
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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 \(2:52:52\)

 

Thanks! :)

 Jul 21, 2024
edited by NotThatSmart  Jul 21, 2024

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