SOLUTION: At 1 pm, ship a leaves port heading due west at x miles per hour. two hours later ship B is 100 miles due south of the same port and heading due north at y miles per hour. at 5 pm

Question 828291: At 1 pm, ship a leaves port heading due west at x miles per hour. two hours later ship B is 100 miles due south of the same port and heading due north at y miles per hour. at 5 pm how far apart are the ship.
Actually i could not understand the meaning of this problem...please elaborate it..
Found 2 solutions by TimothyLamb, richwmiller:
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
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the problem tries to confuse you, but it's basically describing a right triangle, and asking you to find the hypotenuse, given the two legs.
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find the position of both ships at 5 pm so we know the length of legs a and b:
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speed and distance formulas:
s = d/t
d = t*s
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ship1 at 5 pm:
time = 5 - 1 = 4 hours
speed x m/hr
position = 4x west of harbor
a = 4x
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ship2 at 5 pm:
time = 5 - 3 = 2 hours
speed y m/hr
position = (100-2y) south of harbor
b = (100-2y)
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the positions of the ships are perpendicular relative to the harbor, so the ship positions form the legs of the right triangle, and the harbor is the 90 degree vertex.
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hyp = sqrt( aa + bb )
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answer:
hyp = sqrt( 16xx + (100-2y)(100-2y) )
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Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
we are dealing with right triangles here.
we don't have real number speeds just x and y.
4x is the distance traveled west by ship a
3y-100 is the distance traveled north by ship b
the distance between them is the hypotenuse
(4x)^2+(3y-100)^2=d^2
16 x^2+9 y^2-600 y+10000 = d^2

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