At noon, ship A is 140 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

We are given the positions and velocities of two ships, and we are asked to find the rate at which the distance between them is changing at 4:00 PM. Let's define the variables and formulate the problem.

### Step 1: Set up the coordinate system
Let:
- \( A(t) \) be the position of ship A at time \( t \).
- \( B(t) \) be the position of ship B at time \( t \).

At noon (time \( t = 0 \)):
- Ship A is 140 km west of ship B, so the initial position of ship A is \( A(0) = -140 \) km (since it's west of ship B).
- Ship B is at the origin, \( B(0) = 0 \).

### Step 2: Define variables
Let:
- \( x(t) \) be the east-west distance between the ships at time \( t \).
- \( y(t) \) be the north-south distance between the ships at time \( t \).
- \( z(t) \) be the straight-line distance between the ships at time \( t \), which we need to calculate how fast it's changing.

### Step 3: Time-dependent positions
At any time \( t \):
- The position of ship A along the east-west direction (denoted by \( x(t) \)) changes as it moves east at 25 km/h. Since it starts 140 km west of ship B, we have:
  \[
  x(t) = -140 + 25t
  \]
- The position of ship B along the north-south direction (denoted by \( y(t) \)) changes as it moves north at 20 km/h, starting from zero:
  \[
  y(t) = 20t
  \]

### Step 4: Expression for the distance between the ships
The straight-line distance \( z(t) \) between the two ships is given by the Pythagorean theorem:
\[
z(t) = \sqrt{x(t)^2 + y(t)^2}
\]

### Step 5: Differentiate to find the rate of change
We need to find \( \frac{dz}{dt} \) at \( t = 4 \) hours. To do this, differentiate both sides of the equation with respect to \( t \):
\[
\frac{dz}{dt} = \frac{x(t) \frac{dx}{dt} + y(t) \frac{dy}{dt}}{\sqrt{x(t)^2 + y(t)^2}}
\]

We already know:
- \( \frac{dx}{dt} = 25 \) km/h (speed of ship A).
- \( \frac{dy}{dt} = 20 \) km/h (speed of ship B).

### Step 6: Evaluate at \( t = 4 \)
At \( t = 4 \) hours:
- \( x(4) = -140 + 25(4) = -140 + 100 = -40 \) km.
- \( y(4) = 20(4) = 80 \) km.
- The distance \( z(4) \) is:
  \[
  z(4) = \sqrt{(-40)^2 + 80^2} = \sqrt{1600 + 6400} = \sqrt{8000} = 20\sqrt{10} \text{ km}.
  \]

Now, substitute into the derivative expression:
\[
\frac{dz}{dt} = \frac{(-40)(25) + (80)(20)}{20\sqrt{10}} = \frac{-1000 + 1600}{20\sqrt{10}} = \frac{600}{20\sqrt{10}} = \frac{30}{\sqrt{10}} = 3\sqrt{10} \text{ km/h}.
\]

Thus, the rate at which the distance between the ships is changing at 4:00 PM is:
\[
\frac{dz}{dt} = 3\sqrt{10} \approx 9.49 \text{ km/h}.
\]

### Final Answer:
At 4:00 PM, the distance between the ships is changing at approximately \( 9.49 \) km/h.

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Would you like more details on any step? Here are 5 related questions you might be interested in:
1. How does implicit differentiation help in related rates problems?
2. Can this problem be solved using vectors?
3. How does the speed of one ship influence the overall rate of change of distance?
4. What are some real-world applications of related rates problems in navigation?
5. How does changing the ships' speeds affect the outcome?

**Tip**: Always keep track of units when solving related rates problems to ensure consistency in your calculations!

Solve a related rates problem involving two ships, one moving east at 25 km/h and the other moving north at 20 km/h. Learn how to apply the Pythagorean theorem and differentiation to calculate the rate of change in the distance between the ships at 4:00 PM. Detailed solution with key calculus concepts, suitable for high school and early college students.

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