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.)
Well, this is an interesting situation! Let's take a look at it. So, the boat is being pulled into the dock by a winch, and the winch is pulling in rope at a rate of 5 feet per second. We want to find the acceleration of the boat when there is a total of 10 feet of rope out.
Now, here's the thing: the rope is being pulled in at a constant rate of 5 feet per second. This means that the length of the rope is changing, but at a constant rate. And guess what? If a variable is changing at a constant rate, its acceleration is zero!
So, in this case, since the rope is being pulled in at a constant rate of 5 feet per second, the boat's acceleration is zero. It's not speeding up and it's not slowing down.
Now, if only we could find a way to make time go backwards, then we could say the boat is actually decelerating... but I'm afraid that's not an option. So, the acceleration of the boat is simply zero.
Now, wasn't that a rope-tastic answer? Keep those boat-related questions coming!
To find the acceleration of the boat, we need to determine the rate at which the length of the rope is changing.
Let's assume that the length of rope out is represented by the variable l (in feet). We are given that the rate at which the rope is being pulled in is 5 feet per second. This means that the derivative of l with respect to time (t) is 5 ft/s.
We are also given that there is a total of 10 feet of rope out, so we can set up the equation:
l = 10
Now, let's take the derivative of both sides of the equation with respect to time:
d/dt (l) = d/dt (10)
The left side represents the rate at which the length of the rope is changing, which is the acceleration of the boat, denoted by a. The right side is the derivative of a constant, which is zero.
Therefore, we have:
d/dt (l) = 0
Now, we can substitute the given value of the rate of rope being pulled in:
5 ft/s = 0
Hence, the acceleration of the boat is zero.
To find the acceleration of the boat, we need to determine how fast the length of the rope is changing with respect to time.
Given that the winch is pulling in rope at a rate of 5 feet per second, and there is a total of 10 feet of rope out, we can set up the following relationship:
Length of rope = distance from winch to dock - distance from winch to boat
Let's assign variables to the distances. Let x represent the distance from the winch to the dock, and y represent the distance from the winch to the boat.
We know that the winch is 8 feet above the deck of the boat, so we can write:
y = x + 8
Now, let's differentiate both sides of the equation with respect to time (t):
dy/dt = dx/dt
Since the winch pulls in rope at a rate of 5 feet per second, we have:
dy/dt = 5
Now, we substitute the relationship y = x + 8 into the equation dy/dt = 5:
dx/dt = 5
So, the rate at which the distance from the winch to the dock is changing with respect to time is 5 feet per second, which is the acceleration of the boat.
Therefore, the acceleration of the boat is 5 feet per second.
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