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To solve this, we can model the problem as a right triangle where the legs are the distances traveled by each car, and the hypotenuse represents the distance between the two cars. Here’s the solution approach:

### Step 1: Define Variables
- Let \( x \) represent the distance traveled by the car moving south at 48 mi/h.
- Let \( y \) represent the distance traveled by the car moving west at 20 mi/h.
- Let \( z \) represent the distance between the two cars (hypotenuse of the triangle).

### Step 2: Distances Traveled After 2 Hours
Since the cars have been traveling for 2 hours:
- \( x = 48 \times 2 = 96 \) miles (south)
- \( y = 20 \times 2 = 40 \) miles (west)

### Step 3: Distance Between the Cars (Using the Pythagorean Theorem)
\[
z = \sqrt{x^2 + y^2} = \sqrt{96^2 + 40^2} = \sqrt{9216 + 1600} = \sqrt{10816} = 104 \text{ miles}
\]

### Step 4: Rate at Which the Distance Between the Cars is Increasing
We use related rates here. The rate of change of \( z \) with respect to time, \( \frac{dz}{dt} \), is what we need to find.

Since \( z^2 = x^2 + y^2 \), we differentiate both sides with respect to \( t \):
\[
2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
\]
Dividing both sides by \( 2z \):
\[
\frac{dz}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z}
\]

### Step 5: Substitute Values
- \( x = 96 \), \( y = 40 \), \( z = 104 \)
- \( \frac{dx}{dt} = 48 \) mi/h, \( \frac{dy}{dt} = 20 \) mi/h

\[
\frac{dz}{dt} = \frac{(96)(48) + (40)(20)}{104} = \frac{4608 + 800}{104} = \frac{5408}{104} = 52 \text{ mi/h}
\]

### Answer
The rate at which the distance between the cars is increasing after 2 hours is **52 mi/h**.

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Would you like further details or have any questions?

Here are five additional questions to deepen understanding:
1. How would the solution change if one of the cars changed its speed?
2. How can we determine the exact distance between the cars at any given time?
3. What if the cars started at different points instead of the same one?
4. How does using the Pythagorean Theorem help in right triangle distance problems?
5. Why is differentiation necessary to find the rate of distance change?

**Tip:** Related rates problems often require implicit differentiation and understanding relationships between distances and rates.

Two cars start moving from the same point. One travels south at 48 mi/h and the other travels west at 20 mi/h. At what rate (in mi/h) is the distance between the cars increasing two hours later?

Explore a related rates problem where two cars start from the same point, one traveling south at 48 mi/h and the other west at 20 mi/h. Learn how to calculate the rate at which the distance between them is increasing after two hours using the Pythagorean Theorem and differentiation.

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