A police car is located 40 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 120 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 95 feet per second. How fast is the red car actually traveling along the road?

We are given that the distance between a police car and a red car is decreasing at a rate of 95 feet per second, and we need to find how fast the red car is traveling along the road.

### Step-by-Step Explanation:

1. **Setting up the coordinate system:**
   - Let the police car be at a point \((0, 40)\) where 40 feet represents the distance from the road.
   - The red car is driving along the road, so its position on the road can be represented as \((x, 0)\), where \(x\) is the distance of the red car from the police car measured along the road.
   
2. **Using the Pythagorean Theorem:**
   The distance \(d\) between the police car and the red car is the hypotenuse of a right triangle, with legs \(x\) and 40 feet. Therefore, we have:
   \[
   d = \sqrt{x^2 + 40^2}
   \]

3. **Differentiate with respect to time:**
   Since the distance \(d\) is decreasing, we differentiate both sides of the equation with respect to time \(t\):
   \[
   \frac{d}{dt} \left( d \right) = \frac{d}{dt} \left( \sqrt{x^2 + 40^2} \right)
   \]
   Using the chain rule on the right-hand side:
   \[
   \frac{dd}{dt} = \frac{x}{\sqrt{x^2 + 1600}} \cdot \frac{dx}{dt}
   \]
   where \(\frac{dd}{dt}\) is given as \(-95\) ft/sec (negative because the distance is decreasing), and \(\frac{dx}{dt}\) is the speed of the red car along the road (which we need to find).

4. **Substitute known values:**
   At the moment in question, the red car is 120 feet up the road, so \(x = 120\) feet. Also, the police car is 40 feet away from the road, so we can substitute into the equation:
   \[
   -95 = \frac{120}{\sqrt{120^2 + 40^2}} \cdot \frac{dx}{dt}
   \]
   First, calculate the denominator:
   \[
   \sqrt{120^2 + 40^2} = \sqrt{14400 + 1600} = \sqrt{16000} = 126.49 \text{ feet}
   \]
   Now substitute this back into the equation:
   \[
   -95 = \frac{120}{126.49} \cdot \frac{dx}{dt}
   \]
   Simplify:
   \[
   -95 = 0.9487 \cdot \frac{dx}{dt}
   \]
   Solve for \(\frac{dx}{dt}\):
   \[
   \frac{dx}{dt} = \frac{-95}{0.9487} = -100.17 \text{ feet per second}
   \]
   Since speed is always positive, the red car is traveling at approximately **100.17 feet per second** along the road.

### Final Answer:
The red car is traveling at approximately **100.17 feet per second** along the road.

Would you like further details or have any questions? 

Here are 5 related questions to explore:
1. What is the relationship between relative motion and radar measurements?
2. How does the Pythagorean Theorem help in distance-related problems?
3. How does differentiating help find rates of change in distance problems?
4. What real-life factors could affect the accuracy of such radar speed readings?
5. How could you model a similar problem for a curve instead of a straight road?

**Tip:** Always keep track of the signs when working with decreasing or increasing rates in related rates problems.

Learn how to calculate the speed of a red car along a road using related rates and the Pythagorean theorem. The problem involves a police radar measuring a decreasing distance at 95 feet per second, and we find the car's speed as it approaches the police. Suitable for students in Grades 10-12.

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