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This problem describes a situation where an automobile is moving along a road perpendicular to a highway, and we are tasked with determining how fast the distance between the automobile and a farmhouse is increasing when the automobile is 3.2 km past the intersection.

### Given:
- The farmhouse is located \( a = 1.7 \, \text{km} \) from the highway.
- The automobile moves at a speed of \( v = 86 \, \text{km/h} \).
- The distance \( s \) of the automobile from the intersection is 3.2 km at the moment in question.

### Step-by-Step Approach:

1. **Form the relationship using the Pythagorean theorem:**
   - The distance \( l \) between the automobile and the farmhouse forms the hypotenuse of a right triangle, with sides \( a \) (distance from the farmhouse to the road) and \( s \) (distance from the intersection to the automobile).
   - Therefore, the relationship between these variables is:
     \[
     l^2 = a^2 + s^2
     \]
     Substituting \( a = 1.7 \, \text{km} \), we get:
     \[
     l^2 = 1.7^2 + s^2 = 2.89 + s^2
     \]

2. **Differentiate with respect to time \( t \):**
   - To find how fast the distance \( l \) is changing, we differentiate both sides of the equation with respect to \( t \):
     \[
     2l \frac{dl}{dt} = 2s \frac{ds}{dt}
     \]
     Simplifying this expression:
     \[
     l \frac{dl}{dt} = s \frac{ds}{dt}
     \]
   - We know \( \frac{ds}{dt} = 86 \, \text{km/h} \), as this is the speed of the car.

3. **Solve for \( \frac{dl}{dt} \):**
   - First, solve for \( l \) when \( s = 3.2 \, \text{km} \):
     \[
     l = \sqrt{1.7^2 + 3.2^2} = \sqrt{2.89 + 10.24} = \sqrt{13.13} = 3.625 \, \text{km}
     \]
   - Now, substitute into the equation:
     \[
     3.625 \frac{dl}{dt} = 3.2 \times 86
     \]
     \[
     \frac{dl}{dt} = \frac{3.2 \times 86}{3.625} = \frac{275.2}{3.625} = 75.929 \, \text{km/h}
     \]

### Final Answer:
The distance between the automobile and the farmhouse is increasing at approximately **75.929 km/h** when the car is 3.2 km past the intersection.

Let me know if you would like further clarifications or explanations!

Here are five related questions for further exploration:
1. How would the result change if the farmhouse were located 2 km away from the highway?
2. What would happen if the car traveled at a variable speed?
3. How does the speed of the automobile affect the rate at which the distance to the farmhouse changes?
4. Can you derive a general formula for \( \frac{dl}{dt} \) in terms of \( a \), \( s \), and \( v \)?
5. What is the physical interpretation of the instantaneous rate of change \( \frac{dl}{dt} \)?

**Tip:** In problems involving right triangles, always check if the Pythagorean theorem applies—it often simplifies the solution!

An automobile travels past a farmhouse located a = 1.7 km away. The automobile travels at a speed of v = 86 km/h. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 3.2 km past the intersection of the highway and the road?

Learn how to calculate the rate at which the distance between a car moving at 86 km/h and a farmhouse 1.7 km away from the highway is increasing when the car is 3.2 km past the intersection. This problem involves the Pythagorean theorem and related rates in calculus.

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