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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?

Oct 28, 2023

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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?

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

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Oct 28, 2023