math

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?

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asked by Robyn
  1. 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.

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    posted by Chica
  2. 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.

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    posted by Reiny

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