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January 29, 2015

January 29, 2015

Posted by **Anonymous** on Monday, November 30, 2009 at 2:11am.

- algebra -
**drwls**, Monday, November 30, 2009 at 7:18amFor the path-crossing point, you don't need velocities. Let Y be miles north and X = miles east. Let (0,0) be the coordinates of the cruise ship initially.

Y = 350 - X (for tanker)

Y = (5/2) X (for cruise ship)

The cross when

0 = 350 - (7/2)X

X = 100 miles

Y = 250 miles

a) The cruise ship is sqrt [(100)^2 + (250)^2] = 269.3 miles from that point initially. The tanker is

sqrt [(100)^2 + (100)^2] = 141.4 miles away.

b) First calculate when the cruise ship arrives at the crossing point, based upon its known speed:

T = 269.3/40 = 6.73 hours

Then calculate the speed that would put the tanker at that point at that time

V = 141.4 miles/T

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