a police car traveling south towards iowa at 160km/h pursues a truck traveling east wayat 140km/h. at time t=0, the police car is 60km north and the truck is 50km east of iowa. calculate the rate at which the distance between the vehicle is changing at t=10 minutes.

To find the rate at which the distance between the police car and the truck is changing at \( t = 10 \) minutes, we will:

1. Set up the coordinate positions of both vehicles.
2. Express the distance between the vehicles as a function of time.
3. Use the chain rule to differentiate the distance function with respect to time.
4. Evaluate the derivative at \( t = 10 \) minutes.

### Step 1: Setting up positions
At time \( t = 0 \), the police car is 60 km north and the truck is 50 km east of Iowa. Let Iowa be the origin \((0, 0)\) on a coordinate plane.

- **Position of the police car:** Initially at \((0, 60)\). Traveling south at 160 km/h, the position after \( t \) hours is:
  \[
  y_{\text{police}}(t) = 60 - 160t
  \]

- **Position of the truck:** Initially at \((50, 0)\). Traveling east at 140 km/h, the position after \( t \) hours is:
  \[
  x_{\text{truck}}(t) = 50 + 140t
  \]

### Step 2: Express the distance between the vehicles
The distance between the vehicles at time \( t \), \( D(t) \), is given by:
\[
D(t) = \sqrt{[x_{\text{truck}}(t) - 0]^2 + [y_{\text{police}}(t) - 0]^2}
\]
Substitute the expressions for \( x_{\text{truck}}(t) \) and \( y_{\text{police}}(t) \):
\[
D(t) = \sqrt{(50 + 140t)^2 + (60 - 160t)^2}
\]

### Step 3: Differentiate \( D(t) \) with respect to \( t \)
Let’s differentiate \( D(t) \) using the chain rule:
\[
\frac{dD}{dt} = \frac{1}{2\sqrt{(50 + 140t)^2 + (60 - 160t)^2}} \cdot \left[2(50 + 140t) \cdot 140 + 2(60 - 160t)(-160)\right]
\]
This simplifies to:
\[
\frac{dD}{dt} = \frac{(50 + 140t) \cdot 140 + (60 - 160t)(-160)}{\sqrt{(50 + 140t)^2 + (60 - 160t)^2}}
\]

### Step 4: Evaluate \( \frac{dD}{dt} \) at \( t = 10 \) minutes
Convert 10 minutes to hours: \( t = \frac{10}{60} = \frac{1}{6} \) hours.

- Calculate \( x_{\text{truck}}\left(\frac{1}{6}\right) \):
  \[
  x_{\text{truck}}\left(\frac{1}{6}\right) = 50 + 140 \left(\frac{1}{6}\right) = 50 + \frac{140}{6} = 50 + 23.33 = 73.33 \text{ km}
  \]

- Calculate \( y_{\text{police}}\left(\frac{1}{6}\right) \):
  \[
  y_{\text{police}}\left(\frac{1}{6}\right) = 60 - 160 \left(\frac{1}{6}\right) = 60 - \frac{160}{6} = 60 - 26.67 = 33.33 \text{ km}
  \]

Now, compute \( D\left(\frac{1}{6}\right) \):
\[
D\left(\frac{1}{6}\right) = \sqrt{(73.33)^2 + (33.33)^2} = \sqrt{5377.78 + 1111.11} = \sqrt{6488.89} \approx 80.6 \text{ km}
\]

Finally, compute \( \frac{dD}{dt} \left(\frac{1}{6}\right) \):
\[
\frac{dD}{dt} \left(\frac{1}{6}\right) = \frac{(73.33 \cdot 140) + (33.33 \cdot -160)}{80.6}
\]
\[
= \frac{10266.2 - 5332.8}{80.6} = \frac{4933.4}{80.6} \approx 61.2 \text{ km/h}
\]

Thus, the rate at which the distance between the vehicles is changing at \( t = 10 \) minutes is approximately \( 61.2 \text{ km/h} \).

Would you like more details or have any questions? Here are five related questions you might explore:

1. What is the distance between the vehicles at \( t = 0 \)?
2. How would the solution change if the speeds of the vehicles were different?
3. What if the truck were traveling west instead of east?
4. Can you find the time when the vehicles are closest to each other?
5. How does the relative velocity of the vehicles affect the rate of change of distance?

**Tip:** When solving related rates problems, always start by identifying the variables and expressing the relationship between them before differentiating.

A police car traveling south towards Iowa at 160 km/h pursues a truck traveling east at 140 km/h. At time t = 0, the police car is 60 km north and the truck is 50 km east of Iowa. Calculate the rate at which the distance between the vehicles is changing at t = 10 minutes.

This problem involves calculating the rate of change of distance between a police car and a truck using related rates. Given the vehicles' speeds and initial positions, the solution applies the Pythagorean theorem and the chain rule to find the rate of change after 10 minutes.

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