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July 24, 2016
Posted by **Robyn** on Thursday, May 19, 2011 at 11:33pm.

- math -
**Chica**, Friday, May 20, 2011 at 12:01amWell write it out, let 1 represent 10:00 in the x coordinant's position, and let y be the miles out on the course the ship is.

Ship 1

10:00 11:00 12:00 1:00 2:00

(1,10);(2,20);(3,30);(4,40);(5,50);

3:00 4:00 5:00 6:00 7:00

(6,60);(7,70);(8,80);(9,90);(10,100)

Ship 2

1:00 2:00 3:00 4:00 5:00

(4,15);(5,30);(6,45);(7,60);(8,75);

6:00 7:00

(9,90);(10,105)

There is probably an easier way to find it out, though. - math -
**Reiny**, Friday, May 20, 2011 at 8:34amLet t hours be the time since noon, when the second ship passes the first.

distance covered by slower ship = 10(t+3) mile

distance covered by faster ship = 15 t

but at that moment they both had gone the same distance, so ....

15t = 10(t+3)

15t = 10t + 30

5t=30

t = 6

so the second ship passed the first 6 hours past noon or 6:00 pm

check:

slower ship sailed for 9 hours at 10 mph = 90 miles

faster ship sailed for 6 hours at 15 mph = 90 miles

Answer is correct.