To solve this problem, we can apply related rates in calculus. Let’s break it down step by step:

Step 1: Define variables

• Let $x$ represent the horizontal distance between the car and the point on the road directly beneath the plane.
• Let $s$ represent the line-of-sight distance between the plane and the car.
• Let $h = 4$ mi, the height of the plane above the road.
• Let the speed of the plane be $190$ mph.
• Let the rate at which the line-of-sight distance is decreasing be $-165$ mph (since the distance is decreasing, the rate is negative).

Step 2: Relationship between variables

From the geometry of the situation, we can use the Pythagorean theorem to relate the distances: $s^2 = x^2 + h^2$ Substituting $h = 4$ mi: $s^2 = x^2 + 4^2 = x^2 + 16$

Step 3: Differentiate with respect to time $t$

We need to differentiate both sides of the equation with respect to time $t$: $2s \frac{ds}{dt} = 2x \frac{dx}{dt}$ Simplifying: $s \frac{ds}{dt} = x \frac{dx}{dt}$ Where:

• $\frac{ds}{dt}$ is the rate of change of the line-of-sight distance, given as $-165$ mph.
• $\frac{dx}{dt}$ is the car’s speed along the highway, which we are trying to find.
• $s = 5$ mi, as given in the problem.

Step 4: Find $x$

At the instant when $s = 5$ mi, we can use the Pythagorean theorem to solve for $x$: $s^2 = x^2 + 16$ Substituting $s = 5$: $5^2 = x^2 + 16$ $25 = x^2 + 16$ $x^2 = 9$ $x = 3 \text{ mi}$

Step 5: Solve for the car’s speed

Now we substitute the known values into the differentiated equation: $s \frac{ds}{dt} = x \frac{dx}{dt}$ $5 \times (-165) = 3 \times \frac{dx}{dt}$ $-825 = 3 \times \frac{dx}{dt}$ $\frac{dx}{dt} = \frac{-825}{3} = -275 \text{ mph}$ Since speed is the magnitude of velocity, the car’s speed is: $\boxed{275 \text{ mph}}$

Final Answer:

The car's speed along the highway is 275 miles per hour.

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Would you like more details on any of the steps, or do you have any other questions?

Here are 5 related questions for further exploration:

1. How would the car's speed change if the plane's altitude increased to 5 miles?
2. What if the plane was flying at 250 mph instead of 190 mph? How would that affect the car's speed?
3. How does the rate of decrease of the line-of-sight distance change as the car moves closer to the plane?
4. Can you derive the same result using a different method, such as vector analysis?
5. What would happen to the car's speed if the radar measured an increasing line-of-sight distance?

Tip: When solving related rates problems, carefully track how each variable changes with time, and make sure to differentiate correctly!