+0  
 
0
17
1
avatar+1679 

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?

 Oct 28, 2023
 #1
avatar+818 
+1

 

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  

.

 Oct 28, 2023

1 Online Users

avatar