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Let's break down the problem and derive the equation step by step:

### Given:
- Point \( O \) is the intersection of two roads that meet at right angles. One road runs from north to south, and the other from east to west.
- **Car A** is initially 100 meters due west of \( O \) and is moving eastward at \( 20 \, \text{m/s} \).
- **Car B** is initially 80 meters due north of \( O \) and is moving southward at \( 20 \, \text{m/s} \).

We need to show that the square of the distance \( d \) between the two cars after \( t \) seconds is given by:

\[
d^2 = (100 - 20t)^2 + (80 - 20t)^2
\]

### Solution:

1. **Position of Car A after \( t \) seconds**:
   - Car A starts 100 meters west of \( O \), and it is moving east at \( 20 \, \text{m/s} \).
   - After \( t \) seconds, Car A will have moved a distance of \( 20t \) meters to the east. So, its position relative to point \( O \) is:
     \[
     \text{Position of Car A} = 100 - 20t \, \text{meters} \quad \text{(west to east direction)}
     \]

2. **Position of Car B after \( t \) seconds**:
   - Car B starts 80 meters north of \( O \), and it is moving south at \( 20 \, \text{m/s} \).
   - After \( t \) seconds, Car B will have moved a distance of \( 20t \) meters to the south. So, its position relative to point \( O \) is:
     \[
     \text{Position of Car B} = 80 - 20t \, \text{meters} \quad \text{(north to south direction)}
     \]

3. **Distance between Car A and Car B after \( t \) seconds**:
   The two cars are moving along perpendicular roads (east-west and north-south). Therefore, we can use the Pythagorean theorem to calculate the distance \( d \) between them:

   \[
   d^2 = (\text{Position of Car A})^2 + (\text{Position of Car B})^2
   \]
   Substituting the positions of Car A and Car B:
   \[
   d^2 = (100 - 20t)^2 + (80 - 20t)^2
   \]

This matches the expression we were asked to derive.

---

Would you like to go through any part of this process in more detail? Here are five questions that expand on this topic:

1. How would the equation change if the speeds of the cars were different?
2. What is the distance between the cars after 3 seconds?
3. How does the relative velocity between the two cars affect their distance over time?
4. Can you find the minimum distance between the two cars? If so, at what time does it occur?
5. What if Car A was moving at a different speed than Car B—how would you modify the equation?

### Tip:
Always start by defining your variables and understanding how the initial conditions change with time before applying any mathematical formulas.

Point O is the intersection of two roads which cross at right angles; one road runs from north to south, the other from east to west. Car A is 100 metres due west of O and travelling east at a speed of 20 m/s, and Car B is 80 metres due north of O and travelling south at 20 m/s. Show that after t seconds, their distance apart, d metres, is given by d^2 = (100 - 20t)^2 + (80 - 20t)^2.

Explore how to calculate the distance between two cars moving along perpendicular roads using kinematic equations and the Pythagorean theorem. This problem involves deriving an equation to express the distance apart in terms of time and velocity.

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