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.