A highway patrol plane flies 4 mi above a level, straight road at a steady 190 mph. The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 mi, the line-of-sight distance is decreasing at the rate of 165 mph. Find the car's speed along the highway.

Car's speed is_____ miles per hour

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**.

---

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!

A highway patrol plane flies 4 mi above a level, straight road at a steady 190 mph. The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 mi, the line-of-sight distance is decreasing at the rate of 165 mph. Find the car's speed along the highway. Car's speed is_____ miles per hour

Find out how to calculate the speed of a car using related rates and the Pythagorean theorem. This step-by-step solution explains how a highway patrol plane flying at 190 mph and 4 miles above the road measures the car's speed when the line-of-sight distance is decreasing at 165 mph. Ideal for students in Grades 11-12 or college calculus.

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