Step 1: Draw the first displacement of 50 km north.Step 2: From the endpoint of the first displacement, draw the second displacement of 40 km east.Step 3: From the endpoint of the second displacement, draw the third displacement of 30 km south.Net displacement in the North-South direction:
( 50 , \text{km (north)} - 30 , \text{km (south)} = 20 , \text{km (north)} )Net displacement in the East-West direction:
( 40 , \text{km (east)} )Now, the resultant displacement ( R ) can be found by applying the Pythagorean theorem: [ R = \sqrt{(20 , \text{km})^2 + (40 , \text{km})^2} ] [ R = \sqrt{400 , \text{km}^2 + 1600 , \text{km}^2} ] [ R = \sqrt{2000 , \text{km}^2} ] [ R = \sqrt{2000} , \text{km} \approx 44.72 , \text{km} ]The direction of the resultant displacement can be found using the tangent function: [ \theta = \tan^{-1} \left(\frac{20}{40}\right) = \tan^{-1} \left(\frac{1}{2}\right) \approx 26.57^\circ ] The direction is measured from the east towards the north.

Let's break down the problem step by step to ensure everything is clear.

### Displacement Vectors:
1. **First Displacement:**
   - Magnitude: 50 km
   - Direction: North

2. **Second Displacement:**
   - Magnitude: 40 km
   - Direction: East

3. **Third Displacement:**
   - Magnitude: 30 km
   - Direction: South

### Net Displacement in the North-South Direction:
- The first displacement is 50 km North.
- The third displacement is 30 km South.
- Net displacement in the North-South direction:
  \[
  50 \, \text{km (North)} - 30 \, \text{km (South)} = 20 \, \text{km (North)}
  \]

### Net Displacement in the East-West Direction:
- The second displacement is 40 km East.
- There is no displacement in the West direction, so:
  \[
  40 \, \text{km (East)}
  \]

### Resultant Displacement \( R \):
Using the Pythagorean theorem:
\[
R = \sqrt{(20 \, \text{km})^2 + (40 \, \text{km})^2}
\]
\[
R = \sqrt{400 \, \text{km}^2 + 1600 \, \text{km}^2}
\]
\[
R = \sqrt{2000 \, \text{km}^2}
\]
\[
R = \sqrt{2000} \, \text{km} \approx 44.72 \, \text{km}
\]

### Direction of the Resultant Displacement \( \theta \):
The direction can be found using the tangent function:
\[
\theta = \tan^{-1} \left(\frac{20 \, \text{km}}{40 \, \text{km}}\right) = \tan^{-1} \left(\frac{1}{2}\right) \approx 26.57^\circ
\]
The angle \( \theta \) is measured from the east towards the north.

### Final Result:
- **Magnitude of Resultant Displacement:** Approximately 44.72 km.
- **Direction of Resultant Displacement:** 26.57° north of east.

Would you like further details or have any questions? 

Here are some related questions you might find interesting:
1. How would the resultant displacement change if the second displacement was 50 km east instead of 40 km?
2. What if the third displacement was 30 km west instead of south? How would that affect the result?
3. Can you calculate the resultant displacement if the second displacement was in a northwest direction?
4. What if the first displacement was in a northeast direction instead of directly north?
5. How would the direction change if the net displacement in the North-South direction was 40 km instead of 20 km?
6. What is the significance of using the tangent function to find the direction of the resultant displacement?
7. What would the resultant displacement be if the first and third displacements were equal in magnitude and opposite in direction?
8. How would you calculate the resultant displacement if the displacements were given in a 3D space?

**Tip:** When dealing with vector addition in physics, always break the vectors down into their components and use trigonometric functions to calculate the resultant magnitude and direction.

Learn how to calculate resultant displacement step-by-step using vector addition and trigonometric functions. This problem involves vector magnitudes in kilometers and includes the application of the Pythagorean theorem and tangent function to determine both magnitude and direction. Suitable for high school students.

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