A police car is located 30 feet to the side of a straight road.

A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 75 feet per second. How fast is the red car actually traveling along the road?

The actual speed (along the road) of the red car is  feet per second

To solve this problem, we can use related rates in calculus. The problem involves a right triangle where the police car is at one vertex, the red car is on the road (the base of the triangle), and the distance between them forms the hypotenuse. We're given the rate at which the distance between the police car and the red car is decreasing and need to find the red car's speed along the road.

### Step 1: Set up the problem
Let:
- \( y \) be the distance of the red car from the point on the road nearest to the police car.
- \( x \) be the horizontal distance of the red car from the police car, which is the hypotenuse.
- \( z \) be the distance between the red car and the police car, along the road.

We know:
- \( y = 130 \) feet (the distance of the red car along the road).
- \( z = 30 \) feet (the perpendicular distance from the police car to the road).
- \(\frac{dx}{dt} = -75\) feet per second (since the distance between them is decreasing).
- We want to find \(\frac{dy}{dt}\), the speed of the red car along the road.

### Step 2: Relate the variables using the Pythagorean Theorem
\[
x^2 = y^2 + z^2
\]
Differentiating both sides with respect to time \(t\):
\[
2x \frac{dx}{dt} = 2y \frac{dy}{dt}
\]

Simplify:
\[
x \frac{dx}{dt} = y \frac{dy}{dt}
\]

### Step 3: Substitute known values
First, calculate \(x\) using the Pythagorean theorem:
\[
x = \sqrt{y^2 + z^2} = \sqrt{130^2 + 30^2} = \sqrt{16900 + 900} = \sqrt{17800} = \sqrt{17800} \approx 133.5 \text{ feet}
\]

Now substitute \(x \approx 133.5\) feet, \(\frac{dx}{dt} = -75\) feet/second, and \(y = 130\) feet into the differentiated equation:
\[
133.5 \times (-75) = 130 \times \frac{dy}{dt}
\]

### Step 4: Solve for \(\frac{dy}{dt}\)
\[
-10012.5 = 130 \times \frac{dy}{dt}
\]
\[
\frac{dy}{dt} = \frac{-10012.5}{130} \approx -77.02 \text{ feet/second}
\]

Since \(\frac{dy}{dt}\) represents the speed of the red car (a positive quantity):
\[
\boxed{77.02} \text{ feet/second}
\]

Would you like more details or have any questions?

Here are some related questions you might be interested in:
1. How does related rates apply to real-world situations?
2. Can you provide more examples of related rates problems involving triangles?
3. What are some strategies for solving related rates problems?
4. How does the Pythagorean Theorem help in related rates problems?
5. What is the significance of the negative sign in related rates problems?
6. Can related rates problems be solved using parametric equations?
7. How can calculus be applied to motion problems involving vehicles?
8. What is the role of implicit differentiation in solving related rates problems?

**Tip:** When solving related rates problems, carefully identify which quantities are changing and how they relate to each other. This will guide you in setting up the correct equation to differentiate.

Learn how to calculate the speed of a red car using related rates in calculus. This problem involves the police car, the red car, and their changing distances along a road. Detailed step-by-step solution with the application of the Pythagorean theorem. Suitable for advanced high school students studying calculus.

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