Asked by Sagar
Find the acceleration of the specified object. (Hint: Recall that if a variable is changing at a constant rate, its acceleration is zero.)
A boat is pulled into a dock by means of a winch 8 feet above the deck of the boat (see figure). The winch pulls in rope at a rate of 5 feet per second. Find the acceleration of the boat when there is a total of 10 feet of rope out. (Round your answer to three decimal places.)
A boat is pulled into a dock by means of a winch 8 feet above the deck of the boat (see figure). The winch pulls in rope at a rate of 5 feet per second. Find the acceleration of the boat when there is a total of 10 feet of rope out. (Round your answer to three decimal places.)
Answers
Answered by
Steve
draw a diagram. Here is where implicit differentiation comes in handy.
at the moment in question, the boat's distance x from the dock is 8 ft. The length of rope z is found by
z^2 = x^2+8^2
so
x^2 = z^2-64
x = √(z^2-64)
Now start taking derivatives. The velocity and acceleration of the boat are
x' = z/x z' = zz'/x
x" = [(z'^2 + zz")(x) - (zz')x']/x^2
Plugging in the values
x=6 and z=10 and z' = -5 we get
x' = (10/6)(-5) = -25/3
Since z' is constant, z"=0, so
x" = [(25)(6)-(10(-5))(-25/3)]/36 = -200/7 = -7.41 ft/s^2
at the moment in question, the boat's distance x from the dock is 8 ft. The length of rope z is found by
z^2 = x^2+8^2
so
x^2 = z^2-64
x = √(z^2-64)
Now start taking derivatives. The velocity and acceleration of the boat are
x' = z/x z' = zz'/x
x" = [(z'^2 + zz")(x) - (zz')x']/x^2
Plugging in the values
x=6 and z=10 and z' = -5 we get
x' = (10/6)(-5) = -25/3
Since z' is constant, z"=0, so
x" = [(25)(6)-(10(-5))(-25/3)]/36 = -200/7 = -7.41 ft/s^2
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