This problem involves related rates and can be approached using the Pythagorean Theorem.

Step 1: Define variables

Let:

• $x(t)$ represent the distance of the red car from the point on the road nearest to the police car, measured along the road. This is the horizontal distance along the road.
• $y(t)$ represent the distance between the police car and the red car at any given time.
• The police car is 60 feet from the road, so this is a constant. The distance between the red car and the police car forms a right triangle.

We are given:

• $\frac{dy}{dt} = -70$ feet per second (since the distance is decreasing).
• The current distance along the road $x(t) = 150$ feet.
• The police car's distance from the road is $60$ feet.

Step 2: Apply the Pythagorean Theorem

The relationship between $x(t)$, $y(t)$, and the constant 60-foot distance from the road is: $y^2 = x^2 + 60^2$ So, $y^2 = x^2 + 3600$ Differentiate both sides with respect to time $t$: $2y \frac{dy}{dt} = 2x \frac{dx}{dt}$ Simplifying: $y \frac{dy}{dt} = x \frac{dx}{dt}$ We are asked to find $\frac{dx}{dt}$, which is the speed of the red car along the road.

Step 3: Solve for $\frac{dx}{dt}$

First, find $y$ using the Pythagorean Theorem at the given moment when $x = 150$ feet: $y^2 = 150^2 + 60^2 = 22500 + 3600 = 26100$ So, $y = \sqrt{26100} = 161.52 \text{ feet}$

Now, plug in the known values into the related rates equation: $161.52 \times (-70) = 150 \times \frac{dx}{dt}$ Solving for $\frac{dx}{dt}$: $-11306.4 = 150 \times \frac{dx}{dt}$ $\frac{dx}{dt} = \frac{-11306.4}{150} = -75.38 \text{ feet per second}$ Since speed is positive, the red car is traveling at 75.38 feet per second along the road.

Final Answer:

The red car is traveling at approximately 75.4 feet per second along the road.

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Do you want more details, or have any questions?

Here are 5 related questions:

1. How does the rate of change of the distance between two objects relate to their individual speeds?
2. How is the Pythagorean Theorem used to solve related rates problems?
3. If the red car was moving at a different speed, how would the solution change?
4. What happens to the rate of change of distance as the red car moves closer to the police car?
5. How would the problem change if the police car was moving?

Tip: Always ensure units are consistent when working on related rates problems.