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.)

Let's denote the man's position after \( t \) seconds as \( M(t) \) and the woman's position after \( t \) seconds as \( W(t) \).

The man starts walking north at 3 ft/s from point \( P \). Therefore, after \( t \) seconds, the man has walked a distance of \( 3t \) feet north of \( P \).

The woman starts walking south at 5 ft/s from a point 500 feet east of \( P \). However, she starts 5 minutes later, which is 300 seconds later. Therefore, if \( t \) is the total time elapsed since the man started walking, the woman has been walking for \( t - 300 \) seconds. Hence, after \( t \) seconds, she has walked a distance of \( 5(t - 300) \) feet south of her starting point.

We need to find the rate at which the distance between the two people is increasing 15 minutes after the woman starts walking. This is equivalent to 15 minutes plus the 5-minute delay in her start, so \( t = 20 \) minutes or \( t = 1200 \) seconds.

The man's position \( M(t) \) after 1200 seconds is:
\[ M(1200) = 3 \times 1200 = 3600 \text{ feet north of } P \]

The woman's position \( W(t) \) after \( 1200 \) seconds is:
\[ W(1200) = 5 \times (1200 - 300) = 5 \times 900 = 4500 \text{ feet south of her starting point} \]

The distance between their starting points is 500 feet horizontally (east-west direction).

Let's denote the distance between the man and the woman as \( d(t) \). We can use the Pythagorean theorem to relate \( d(t) \) to their positions:
\[ d(t) = \sqrt{(500)^2 + (M(t) + W(t))^2} \]

For \( t = 1200 \) seconds,
\[ d(1200) = \sqrt{(500)^2 + (3600 + 4500)^2} = \sqrt{500^2 + 8100^2} \]
\[ d(1200) = \sqrt{250000 + 65610000} = \sqrt{65860000} = 8115.42 \text{ feet (approximately)} \]

To find the rate at which they are moving apart, we need to differentiate \( d(t) \) with respect to time \( t \).

Let \( x = 500 \) (constant) and \( y(t) = 3600 + 4500 \text{ at } t = 1200 \).

\[ d(t) = \sqrt{x^2 + y(t)^2} \]

Using the chain rule:
\[ \frac{d}{dt} [d(t)] = \frac{1}{2} [x^2 + y(t)^2]^{-1/2} \cdot 2y(t) \cdot \frac{dy(t)}{dt} \]

Given:
\[ x = 500 \]
\[ y(t) = 3600 + 4500 = 8100 \]
\[ \frac{dy(t)}{dt} = 3 + 5 = 8 \text{ ft/s} \]

Then,
\[ \frac{d}{dt} [d(t)] = \frac{8100}{\sqrt{500^2 + 8100^2}} \cdot 8 \]
\[ = \frac{8100}{8115.42} \cdot 8 \]
\[ = 0.998 \cdot 8 \]
\[ \approx 7.98 \text{ ft/s} \]

Therefore, the rate at which the distance between the man and the woman is increasing 15 minutes after the woman starts walking is approximately \( 7.98 \text{ ft/s} \).

Would you like details or have any questions?

Here are some related questions for further understanding:
1. How is the chain rule applied in differentiation?
2. What is the significance of the Pythagorean theorem in this problem?
3. How can the concept of relative velocity be used in similar problems?
4. What are the practical applications of such rate problems?
5. How would the problem change if the walking speeds were different?
6. What if the woman started walking west instead of south?
7. How does the delay in the woman's start affect the solution?
8. Can this problem be solved using vectors? If so, how?

**Tip:** Always check units and ensure they are consistent throughout the problem to avoid calculation errors.

Explore a detailed solution to a rate of separation problem involving a man and a woman walking at different speeds and directions. Understand how to apply the Pythagorean theorem and differentiation to find the rate at which they are moving apart. Suitable for high school students studying kinematics and calculus.

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