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The problem involves two vehicles: a police car and a truck, moving at different speeds and directions away from Sioux Falls. The police car is traveling south at 160 km/h, and the truck is moving east at 140 km/h. We are asked to find the rate at which the distance between them is changing after 10 minutes.

Let's break down the problem mathematically:

1. **Initial Setup (At \( t = 0 \)):**
   - The police car is 50 km north of Sioux Falls.
   - The truck is 30 km east of Sioux Falls.

   We can use the Pythagorean theorem to represent the distance between the two vehicles as a function of time.

   Let:
   - \( x(t) \) be the eastward distance of the truck from Sioux Falls.
   - \( y(t) \) be the southward distance of the police car from Sioux Falls.
   - \( z(t) \) be the distance between the truck and the police car.

   From the problem, the truck moves east at a speed of 140 km/h, and the police car moves south at a speed of 160 km/h.

2. **Relationships:**
   The distance \( z(t) \) between the vehicles is given by the Pythagorean theorem:
   \[
   z(t) = \sqrt{x(t)^2 + y(t)^2}
   \]
   We need to find how fast \( z(t) \) is changing at \( t = 10 \) minutes.

3. **Rates of Change:**
   The rate of change of \( z \) with respect to time is found using implicit differentiation:
   \[
   \frac{dz}{dt} = \frac{1}{2} \cdot \frac{2x(t) \frac{dx}{dt} + 2y(t) \frac{dy}{dt}}{z(t)}
   \]
   Simplifying:
   \[
   \frac{dz}{dt} = \frac{x(t) \cdot \frac{dx}{dt} + y(t) \cdot \frac{dy}{dt}}{z(t)}
   \]

4. **Specific Values at \( t = 10 \) minutes:**
   At \( t = 10 \) minutes (which is \( \frac{1}{6} \) hours):
   - The truck's eastward position: \( x(10) = 30 + 140 \times \frac{1}{6} = 53.333 \) km.
   - The police car's southward position: \( y(10) = 50 - 160 \times \frac{1}{6} = 23.333 \) km.

   The rate of change of the truck's position is \( \frac{dx}{dt} = 140 \) km/h, and the rate of change of the police car's position is \( \frac{dy}{dt} = 160 \) km/h.

5. **Final Calculation:**
   First, calculate \( z(10) \):
   \[
   z(10) = \sqrt{53.333^2 + 23.333^2} = \sqrt{2844.44 + 544.44} = \sqrt{3388.88} \approx 58.228 \text{ km}.
   \]
   Now, calculate \( \frac{dz}{dt} \):
   \[
   \frac{dz}{dt} = \frac{53.333 \times 140 + 23.333 \times 160}{58.228} = \frac{7466.67 + 3733.33}{58.228} = \frac{11200}{58.228} \approx 192.297 \text{ km/h}.
   \]

Thus, the rate at which the distance between the vehicles is changing at \( t = 10 \) minutes is approximately **192.297 km/h**.

Would you like more details or have any questions?

Here are five related questions:
1. How would the problem change if the speeds of the vehicles were reversed?
2. What if the truck and police car were both moving in the same direction?
3. How does implicit differentiation apply to real-world motion problems?
4. Can you determine at what time the distance between the vehicles is minimal?
5. How does the Pythagorean theorem apply in calculating changing distances in motion?

**Tip:** Always keep track of units (hours, minutes) when calculating speeds and distances to avoid errors in real-world problems.

A police car traveling south toward Sioux Falls, Iowa, at 160 km/h pursues a truck traveling 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 = 10 minutes.

Solve a related rates problem involving a police car and a truck moving away from Sioux Falls. Learn how to use the Pythagorean theorem and implicit differentiation to calculate the rate at which the distance between them is changing after 10 minutes. Suitable for Grade 11-12 students studying Pre-Calculus or Calculus.

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