SOLUTION: Two boats leave the dock at 1:00pm. Each boat travels in a straight line at a constant speed, and the two lines along which the boats are traveling are perpendicular to one another

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Question 1130701: Two boats leave the dock at 1:00pm. Each boat travels in a straight line at a constant speed, and the two lines along which the boats are traveling are perpendicular to one another. At 3:00pm, the boats are 16 miles apart. If the first boat travels 6 miles per hour faster than the second, find the speed of each boat.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Positions and departure point form right triangle.
Two hours of travel time.

r, speed of the second boat
r+6, speed of first boat

%282r%29%5E2%2B%282%28r%2B6%29%29%5E2=16%5E2
Take it from there.

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
If the speed of the slower boat is  "r"  miles per hour, then the speed of the faster boat is  (r+6) miles per hour.


You have the right angle triangle with the legs  2r  and  2*(r+6)  miles.

The hypotenuse is 16 miles, which gives you an equation


(2r)^2 + (2*(r+6))^2 = 16^2


4r^2 + 4r^2 + 48 r + 144 = 256


8r^2 + 48r - 112 = 0

r^2 + 6r - 14 = 0


r%5B1%2C2%5D = %28-6+%2B-+sqrt%286%5E2+%2B+4%2A14%29%29%2F2 = %28-6+%2B-+9.592%29%2F2.


The only meaningful solution is the positive root  r = %28-6+%2B+9.592%29%2F2 = 1.796 miles per hour.


Answer.  1.796 mph for the slower boat and  7.796 mph for the faster boat.


Check.   %282%2A1796%29%5E2+%2B+%282%2A7.796%29%5E2 = 256.0 = 16^2.    ! Correct !

Solved.

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Be aware !.   The equation   r%5E2%2B%28r%2B6%29%5E2=16%5E2   by  @josgarithmetic from his post is   W R O N G.


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O-o-o ! He just re-wrote it correctly from my post !