a boat is pulled into a dock by a rote attached to the bow of the boat and passing through a pulley on the dock that is 1 meter high than the bow of the boat. If the rope is pulled in at a rate of 1 meter per second, how fast is the boat approaching the dock when it is 8 meters from the dock?
2006-12-10 15:12:21 · 2 answers · asked by pago 1 in Science & Mathematics ➔ Physics
Start with the Pythagorean theorem. The distance from the boat to the dock X is one side adjacent to the right angle of a right triangle and the height of the dock attach point Y is the other. The hypotenuse H = sqrt(X^2 + Y^2) = sqrt(8^2 + 1^2). The hypotenuse is decreasing at a rate of 1 m/s. The approach rate is 1*H/X = sqrt(8^2 + 1^2)/8 m/s.
The reason why isn't immediately obvious. Let's call the angle between X and H theta. For a small delta-H times a force F, the end of the rope exerts a force of F*cos(theta) along X. Then conservation of energy says delta-X = delta-H/cos(theta). Thus the rate of change of X = rate of change of H/cos(theta) = 1*H/X.
2006-12-10 15:41:27 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
For every meter that the pully advances the boat advances 7/8ths meters. Therefore it is advancing 7/8th meters/ second
2006-12-10 23:23:06 · answer #2 · answered by eric l 6 · 0⤊ 0⤋