David walks 3km north and then 4km east. What is his distance and displacement?

To find David's **distance** and **displacement**, we approach the problem as follows:

### 1. **Distance:**
The total distance is the sum of all the paths David walked. He walks:
- 3 km north
- 4 km east

Thus, the total distance traveled is:
\[
\text{Distance} = 3 \, \text{km} + 4 \, \text{km} = 7 \, \text{km}
\]

### 2. **Displacement:**
Displacement is the straight-line distance between David's starting and ending points. Since David's movement forms a right-angled triangle (with the northward and eastward walks as the two legs), we can use the **Pythagorean theorem** to find the displacement.

Let the displacement be \( d \). According to the Pythagorean theorem:
\[
d^2 = (3 \, \text{km})^2 + (4 \, \text{km})^2
\]
\[
d^2 = 9 + 16 = 25
\]
\[
d = \sqrt{25} = 5 \, \text{km}
\]

### **Summary:**
- **Distance traveled** = 7 km
- **Displacement** = 5 km in a northeast direction.

Would you like more details on any part of this solution?

Here are five related questions:
1. What would the displacement be if David walked 5 km north and 12 km east?
2. How do you calculate the angle of the displacement vector in this scenario?
3. What is the difference between distance and displacement in general terms?
4. How would you compute displacement for a journey that forms a different triangle?
5. How does the Pythagorean theorem apply to real-world navigation?

**Tip:** Displacement is always a straight line from start to end, regardless of the path taken.

Learn how to calculate David's total distance traveled and his displacement using the Pythagorean theorem. This geometry problem involves a 3 km walk north and a 4 km walk east, making it perfect for students in Grades 6-8. Follow a step-by-step explanation of the solution, including distance and displacement calculations.

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