Posted by Robyn on .
A ship left at 9:00 am going 10 miles per hour on a course. Another ship left at noon on the same course going 15 miles per hour. At what time did the second ship overtake the first ship?

math 
Chica,
Well 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,
Let 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.