Two fishing boats depart a harbour at the same time, one traveling east, the other south. The eastbound boat travels at a speed 3 mi/h faster than the southbound boat.
After two hours the boats are 30 mi apart from each other
Find the speed of the southbound boat.
2007-09-12 03:40:31 · 2 answers · asked by alexo 4 in Science & Mathematics ➔ Mathematics
let s = speed of the southbound boat
s+3 = speed of the eastbound boat
d = distance traveled of the southbound boat = 2s
e = distance traveled of the eastbound boat = 2(s+3) = 2s+6
If you draw a diagram of this you will have a right triangle with the 2 legs being the distances the boats traveled and the hypotenuse being 30 miles.
Using Pythagoras
d^2 + e^2 = 30^2
(2s)^2 + (2s+6)^2 = 900
4s^2 + 4s^2 +24s + 36 = 900
8s^2 + 24s - 864 = 0
s^2 + 3s - 108 = 0
(s-9)(s+12) = 0
s=9 or s=-12.
Since the speed of the boat can not be negative, then:
s = 9 mph
2007-09-12 04:01:14 · answer #1 · answered by T 5 · 0⤊ 0⤋
The southbound boat is travelling at s mi/hour; the eastbound boat is travelling at s+3. After two hours, the southbound boat is 2s miles away; the eastbound boat is 2(s+3) miles away.
Now picture a triangle with sides of length 2s and 2(s+3); they hypotenuse is 30. So...
30^2 = (2s)^2 + [2(s+30]^2, a quadratic equation in s. Solve for s.
2007-09-12 10:51:09 · answer #2 · answered by Mathsorcerer 7 · 0⤊ 0⤋