A man starts walking north at 3 ft/s from a point *P*. Five minutes later a woman starts walking south at 5 ft/s from a point 500 ft due east of *P*. At what rate (in ft/s) are the people moving apart 15 minutes after the woman starts walking? (Round your answer to two decimal places.)

To solve this problem, we'll break it down step-by-step. We'll determine the positions of the man and the woman, their distance from each other, and the rate at which this distance is changing.

Let's define our coordinate system with point \( P \) as the origin \((0,0)\).

### Positions After 15 Minutes
- The man starts walking north at 3 ft/s.
- The woman starts walking south at 5 ft/s, 5 minutes after the man starts walking.

#### Position of the Man
The man starts walking at \( t = 0 \) and continues walking for \( t = 20 \) minutes (since the woman starts 5 minutes later and we are looking at 15 minutes after she starts).

So, the man's position after 20 minutes is:
\[
\text{Northward distance by man} = 3 \text{ ft/s} \times 20 \text{ min} \times 60 \text{ s/min} = 3600 \text{ ft}
\]

Thus, his position is \((0, 3600)\).

#### Position of the Woman
The woman starts walking at \( t = 5 \) minutes and walks for \( t = 15 \) minutes.

So, the woman's position after 15 minutes is:
\[
\text{Southward distance by woman} = 5 \text{ ft/s} \times 15 \text{ min} \times 60 \text{ s/min} = 4500 \text{ ft}
\]

Her starting point is 500 ft east of \( P \), so her position is \((500, -4500)\).

### Distance Between the Man and the Woman
Now, we need to calculate the distance between their positions \((0, 3600)\) and \((500, -4500)\).

\[
\text{Distance} = \sqrt{(500 - 0)^2 + (-4500 - 3600)^2} = \sqrt{500^2 + (-8100)^2} = \sqrt{250000 + 65610000} = \sqrt{65860000}
\]

\[
\text{Distance} = 8114.08 \text{ ft (approx)}
\]

### Rate of Change of Distance
To find the rate at which they are moving apart, we use the related rates concept.

Let's denote:
- \( x \) as the east-west distance, which is constant at 500 ft.
- \( y \) as the north-south distance, which changes over time.

The north-south distance is given by:
\[
y = \text{Man's distance} + \text{Woman's distance} = 3600 + 4500 = 8100 \text{ ft}
\]

We need the rate of change of the distance \( d \) between them with respect to time:
\[
d = \sqrt{x^2 + y^2} = \sqrt{500^2 + 8100^2}
\]

Differentiate \( d \) with respect to time \( t \):
\[
2d \frac{dd}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]

Given \( \frac{dx}{dt} = 0 \) (since the east-west distance is constant),
\[
2d \frac{dd}{dt} = 2y \frac{dy}{dt}
\]

Substitute the known values:
\[
d = 8114.08 \text{ ft}, \quad y = 8100 \text{ ft}, \quad \frac{dy}{dt} = 3 + 5 = 8 \text{ ft/s}
\]

\[
8114.08 \frac{dd}{dt} = 8100 \times 8
\]

\[
\frac{dd}{dt} = \frac{8100 \times 8}{8114.08} = \frac{64800}{8114.08} \approx 7.99 \text{ ft/s}
\]

Thus, the rate at which they are moving apart 15 minutes after the woman starts walking is approximately 7.99 ft/s.

Would you like more details or have any questions?

Here are some related questions:
1. How can you set up a coordinate system for solving motion problems?
2. What is the importance of converting time units consistently?
3. How do you apply the Pythagorean theorem in related rates problems?
4. How do you differentiate distance with respect to time?
5. Why is the east-west distance constant in this problem?
6. How do you interpret the north-south positions of the man and the woman?
7. What are the steps for calculating the rate of change of distance between two moving points?
8. Can this method be applied to other problems involving motion in different directions?

**Tip:** Always clearly define the variables and their relationships when dealing with related rates problems to simplify the differentiation process.

Learn how to solve the problem of calculating rates of motion and distance change between a man and a woman walking in opposite directions. This problem involves coordinate geometry and related rates formulas, suitable for advanced high school students.

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