SOLUTION: Please help me with this problem: Two boats leave the buoy at the same time. One of the boats travels north at a rate of N miles per hour, and the other boat travels east at a rate

Question 990307: Please help me with this problem: Two boats leave the buoy at the same time. One of the boats travels north at a rate of N miles per hour, and the other boat travels east at a rate of 4N miles per hour. What is the distance between the boats 30 minutes after both boats leave the buoy?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Two boats leave the buoy at the same time.
One of the boats travels north at a rate of N miles per hour, and the other boat travels east at a rate of 4N miles per hour.
What is the distance between the boats 30 minutes after both boats leave the buoy?
:
Find the distances they travel in .5 hrs (30 min). Dist = speed * time
North: .5n
East: .5(4n) = 2n
:
This is a pythag problem; a^2 + b^2 = c^2; where
a = .5n
b = 2n
c = the distance between them
c^2 = (.5n)^2 + (2n)^2
c^2 = .25n^2 + 4n^2
like terms
c^2 = 4.25n^2
c =
Factor inside the radical to reveal perfect squares
c =
Extract the square root of those
c = or about 2.06 miles apart after 30 min
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