A police car travelling south toward Sioux Falls, Iowa, at 160 km/h pursues a truck travelling east away from Sioux Falls at 140 km/h. At time t = 0, the police car is 50 km north and the truck is 30 km east of Sioux Falls. Calculate the rate at which the distance between the vehicles is changing at t = 12 minutes. (Use decimal notation. Give your answer to three decimal places.)

To solve the problem, we can model the positions of the police car and the truck relative to Sioux Falls and then apply the Pythagorean theorem to find the rate of change of the distance between them.

1. **Define the positions**:
   - Let \( P(t) \) be the position of the police car.
   - Let \( T(t) \) be the position of the truck.
   - At \( t = 0 \):
     - The police car is at \( (0, 50) \) km (50 km north of Sioux Falls).
     - The truck is at \( (30, 0) \) km (30 km east of Sioux Falls).

2. **Velocity components**:
   - The police car travels south at 160 km/h, which affects its \( y \)-coordinate:
     \[
     y_P(t) = 50 - 160t
     \]
   - The truck travels east at 140 km/h, affecting its \( x \)-coordinate:
     \[
     x_T(t) = 30 + 140t
     \]

3. **Find the positions at \( t = 12 \) minutes** (or \( t = 0.2 \) hours):
   - For the police car:
     \[
     y_P(0.2) = 50 - 160(0.2) = 50 - 32 = 18 \text{ km}
     \]
   - For the truck:
     \[
     x_T(0.2) = 30 + 140(0.2) = 30 + 28 = 58 \text{ km}
     \]

4. **Distance between the vehicles**:
   - The distance \( D \) between the police car and the truck can be expressed as:
     \[
     D = \sqrt{(x_T(t))^2 + (y_P(t))^2}
     \]
   - At \( t = 0.2 \):
     \[
     D = \sqrt{(58)^2 + (18)^2} = \sqrt{3364 + 324} = \sqrt{3688} \approx 60.754 \text{ km}
     \]

5. **Rate of change of distance**:
   - To find the rate at which the distance is changing, we need to use the chain rule:
     \[
     \frac{dD}{dt} = \frac{x_T(t) \frac{dx_T}{dt} + y_P(t) \frac{dy_P}{dt}}{D}
     \]
   - The derivatives:
     - \( \frac{dx_T}{dt} = 140 \) km/h (eastward)
     - \( \frac{dy_P}{dt} = -160 \) km/h (southward)

6. **Substituting values**:
   - At \( t = 0.2 \):
     - \( x_T(0.2) = 58 \)
     - \( y_P(0.2) = 18 \)
     - \( D \approx 60.754 \)

   - Calculate \( \frac{dD}{dt} \):
     \[
     \frac{dD}{dt} = \frac{58(140) + 18(-160)}{60.754}
     \]
     \[
     = \frac{8120 - 2880}{60.754} = \frac{5240}{60.754} \approx 86.275 \text{ km/h}
     \]

Thus, the rate at which the distance between the vehicles is changing at \( t = 12 \) minutes is approximately **86.275 km/h**.

### Summary
- Rate of change of distance: **86.275 km/h**.

Would you like more details or have any questions about this solution? Here are some related questions for further exploration:

1. How would the result change if the truck traveled at a different speed?
2. What would be the positions of the vehicles after 30 minutes?
3. How can we graphically represent the paths of the police car and the truck?
4. What is the maximum distance between the two vehicles during their chase?
5. How does the angle of their paths affect the rate of distance change? 

**Tip**: When dealing with rates of change in relative motion, it's helpful to visualize the scenario with a diagram for clarity.

A police car travelling south toward Sioux Falls, Iowa, at 160 km/h pursues a truck travelling east away from Sioux Falls at 140 km/h. At time t = 0, the police car is 50 km north and the truck is 30 km east of Sioux Falls. Calculate the rate at which the distance between the vehicles is changing at t = 12 minutes.

Discover how to calculate the rate at which the distance between a police car and a truck is changing when they travel at different speeds. This problem involves relative motion and the Pythagorean theorem, yielding a result of approximately 86.275 km/h after 12 minutes.

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