A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat.&nbsp;(a) The winch pulls in rope at a rate of 4 feet per second. Determine the speed of the boat when there is 13 feet of rope out. What happens to the speed of the boat as it gets closer to the dock?&nbsp;(b) Suppose the boat is moving at a cons tant rate of 4 feet per second. Determine the speed at which the winch pulls in rope when the re is a tota l of 13 feet of rope out. What happens to the speed at which the winc h pulls in rope as the boat gets closer to the dock ?

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A boat is pulled into a dock by means of a winch 12 feet above the deck of the boat.&amp;nbsp;(a) The winch pulls in rope at a rate of 4 feet per second. Determine the speed of the boat when there is 13 feet of rope out. What happens to the speed of the boat as it gets closer to the dock?&amp;nbsp;(b) Suppose the boat is moving at a cons tant rate of 4 feet per second. Determine the speed at which the winch pulls in rope when the re is a tota l of 13 feet of rope out. What happens to the speed at which the winc h pulls in rope as the boat gets closer to the dock ?

memomzungup4 · Accepted Answer

Let l be the length of the rope. Then l2=144+x2  (a)2ldldt=2xdxdt, so dxdt=lx⋅dldt=lx⋅−4=−4lx  For l=13,x=5, and dxdt=−4⋅135=−525ftsec⁡  (b)dldt=xldxdt=513⋅−4=−2013ftsec⁡  So as L→0,dldt increases  Therefore, as the boat gets closer to the dock, its speed increases.  Result: (a)−525ftsec⁡, (b)−2013ftsec⁡  As the boat gets closer to the dock, its speed increases.

RizerMix · Accepted Answer

(a) Let x be the distance between the boat and the dock. When there is 13 feet of rope out, we have x=12+13=25 feet. We can use the Pythagorean theorem to relate the distance x to the length L of the rope, which is decreasing at a rate of 4 feet per second:x2+122=L2Taking the derivative of both sides with respect to time t, we get:2xdxdt=2LdLdtSubstituting x=25 and dLdt=−4, we can solve for dxdt, which represents the speed of the boat:dxdt=LxdLdt=252−12225(−4)=−4825&amp;nbsp;ft/sAs the boat gets closer to the dock, the length of the rope decreases, which means the boat&#039;s speed decreases as well.(b) Now suppose the boat is moving at a constant rate of 4 feet per second. We can again use the Pythagorean theorem to relate the distance x to the length L of the rope:x2+122=L2Taking the derivative of both sides with respect to time t, we get:2xdxdt=2LdLdtSubstituting x=25 and dxdt=4, we can solve for dLdt, which represents the speed at which the winch pulls in rope:dLdt=2xLdxdt=2·25252−122·4=2007&amp;nbsp;ft/sAs the boat gets closer to the dock, the length of the rope decreases, which means the winch&#039;s speed also decreases.

Jeffrey Jordon · Accepted Answer

(a) Let d be the distance from the boat to the dock and r be the length of rope out. Then we have a right triangle with hypotenuse of length r, one leg of length d, and the other leg of length 12. Using the Pythagorean theorem, we have r2=d2+122. Differentiating with respect to time, we get 2rdrdt=2ddddt.When r=13, we have 169=d2+144, so d=25=5. Therefore, dddt=−drdrdt=−513(4)=−2013 ft/s.The speed of the boat is given by dddt. As the boat gets closer to the dock, d decreases and dddt becomes more negative, so the boat slows down.(b) Let v be the speed of the boat. Then we have ddt(r2)=ddt(d2+144)=2ddddt=2vr2−144. When r=13, we have ddt(169)=2v132−144, so v=13132−144=135.The speed at which the winch pulls in rope is given by drdt. As the boat gets closer to the dock, r decreases and drdt becomes smaller, so the winch pulls in rope more slowly.

Vasquez · Accepted Answer

(a) Let d be the distance from the boat to the dock and r be the length of rope out. Then, we have r2=d2+122 and 2rdrdt=2ddddt.To find the speed of the boat when r=13, we can use implicit differentiation. Differentiating both sides of r2=d2+144 with respect to time, we get 2rdrdt=2ddddt. At the moment when r=13, we have 169=d2+144, so dddt=−2013 ft/s. Therefore, dddt is the speed of the boat when r=13. As the boat gets closer to the dock, d decreases and dddt becomes more negative, so the boat slows down.(b) Again, we have ddt(r2)=ddt(d2+144)=2ddddt=2vr2−144. To find the speed of the winch when r=13, we can use the chain rule: ddt(r2)=2rdrdt and ddt(d2)=2ddddt. Then we have 2rdrdt=2vr2−144 and ddt(r2)=2rdrdt. At the moment when r=13, we have ddt(169)=2(13)drdt, so drdt=13v132−144=135 ft/s. As the boat gets closer to the dock, r decreases and drdt becomes smaller, so the winch pulls in rope more slowly.

A boat is pulled into a dock by means of a winch 12 feet above the deck of the b

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