We are given the following details:

• At noon, ship A is 40 nautical miles west of ship B.
• Ship A is moving west at 20 knots.
• Ship B is moving north at 16 knots.
• We are to find how fast the distance between the two ships is changing at 5 PM.

Step 1: Define Variables

Let’s set up a coordinate system where:

• Ship B starts at the origin $(0, 0)$, and moves north.
• Ship A starts at the point $(-40, 0)$, 40 nautical miles due west of ship B, and moves west.

Let:

• $x$ be the horizontal position of ship A at any time $t$, where $t$ is the time in hours after noon. Since ship A moves west, $x(t)$ is decreasing.
• $y$ be the vertical position of ship B at any time $t$. Ship B is moving north, so $y(t)$ increases.

We know that:

• Ship A starts at $x(0) = -40$ and moves west at 20 knots, so $\frac{dx}{dt} = -20$ knots.
• Ship B starts at $y(0) = 0$ and moves north at 16 knots, so $\frac{dy}{dt} = 16$ knots.

Step 2: Write an Expression for the Distance Between the Ships

Let $z(t)$ be the distance between the two ships at time $t$. Using the Pythagorean theorem, the distance between the ships at any time $t$ is given by:

$z(t) = \sqrt{(x(t))^2 + (y(t))^2}$

Step 3: Differentiate with Respect to Time

To find the rate at which the distance between the ships is changing, we need to differentiate $z(t)$ with respect to time $t$. Using implicit differentiation:

$\frac{d}{dt} \left( z(t)^2 \right) = \frac{d}{dt} \left( (x(t))^2 + (y(t))^2 \right)$

This gives:

$2z(t) \frac{dz}{dt} = 2x(t) \frac{dx}{dt} + 2y(t) \frac{dy}{dt}$

Simplifying:

$z(t) \frac{dz}{dt} = x(t) \frac{dx}{dt} + y(t) \frac{dy}{dt}$

Now, we will plug in the known values and calculate the rate of change of the distance between the ships at 5 PM (when $t = 5$ hours).

Step 4: Find the Positions of the Ships at 5 PM

At $t = 5$:

• The position of ship A: $x(5) = -40 - 20 \times 5 = -40 - 100 = -140$ nautical miles.
• The position of ship B: $y(5) = 0 + 16 \times 5 = 80$ nautical miles.

Step 5: Calculate the Distance Between the Ships at 5 PM

Now, we can calculate $z(5)$:

$z(5) = \sqrt{(-140)^2 + (80)^2} = \sqrt{19600 + 6400} = \sqrt{26000} \approx 161.24 \text{ nautical miles}$

Step 6: Find the Rate of Change of the Distance Between the Ships

Using the formula for $\frac{dz}{dt}$:

$z(5) \frac{dz}{dt} = x(5) \frac{dx}{dt} + y(5) \frac{dy}{dt}$

Substitute the known values:

$161.24 \times \frac{dz}{dt} = (-140)(-20) + (80)(16)$

$161.24 \times \frac{dz}{dt} = 2800 + 1280 = 4080$

Solve for $\frac{dz}{dt}$:

$\frac{dz}{dt} = \frac{4080}{161.24} \approx 25.3 \text{ knots}$

Final Answer:

The distance between the ships is changing at a rate of approximately 25.3 knots at 5 PM.

Would you like any further clarification or details on this problem? Here are some related questions for further exploration:

1. How would the solution change if both ships were moving at different speeds in the same direction?
2. How does the distance between the ships change if one of them changes direction during the motion?
3. How could we determine the point where the ships will meet if they continue on their current paths?
4. What are the implications of the relative velocities of the ships on the rate of distance change?
5. How can we use vector addition to approach similar problems involving moving objects?

Tip: When dealing with related rates problems like this, always start by identifying the relevant variables, write down the equation that relates them, and differentiate carefully with respect to time.