To reach his school from his house, John walks 200m[E],100m[N],150m[NW] and finally

300[S30\deg W].

[4 marks]

[1] a) Draw a diagram to represent these displacements. [1]

[3] b) Find the magnitude of the total displacement. [3]

### Part [1] a) Drawing a Diagram
To represent the displacements on a diagram, follow these steps:

1. **200 m [E]:** Draw a horizontal line 200 meters to the right (east).
2. **100 m [N]:** From the end of the first line, draw a vertical line 100 meters upwards (north).
3. **150 m [NW]:** From the end of the second line, draw a line at a \( 45^\circ \) angle to the horizontal, 150 meters in the northwest direction. Northwest means the line goes upward and to the left.
4. **300 m [S30°W]:** From the end of the third line, draw a line at a \( 30^\circ \) angle south of west, 300 meters long.

### Part [3] b) Finding the Magnitude of the Total Displacement

To find the total displacement, break down each vector into its components and sum them.

#### Step 1: Break down each vector into x (East-West) and y (North-South) components.

1. **Vector 1: 200 m [E]**
   - \( x_1 = 200 \, \text{m} \)
   - \( y_1 = 0 \, \text{m} \)

2. **Vector 2: 100 m [N]**
   - \( x_2 = 0 \, \text{m} \)
   - \( y_2 = 100 \, \text{m} \)

3. **Vector 3: 150 m [NW]**
   - Since this is in the northwest direction (45° to both the x and y axes):
   \[
   x_3 = 150 \times \cos(45^\circ) = 150 \times \frac{1}{\sqrt{2}} \approx 150 \times 0.707 = -106.1 \, \text{m}
   \]
   \[
   y_3 = 150 \times \sin(45^\circ) = 150 \times \frac{1}{\sqrt{2}} \approx 150 \times 0.707 = 106.1 \, \text{m}
   \]

4. **Vector 4: 300 m [S30°W]**
   - This vector makes a 30° angle with the west direction (or 180° from east):
   \[
   x_4 = 300 \times \cos(30^\circ) = 300 \times \frac{\sqrt{3}}{2} \approx 300 \times 0.866 = -259.8 \, \text{m}
   \]
   \[
   y_4 = 300 \times \sin(30^\circ) = 300 \times \frac{1}{2} = -150 \, \text{m}
   \]

#### Step 2: Sum the components
- **Total x-component:**
  \[
  R_x = x_1 + x_2 + x_3 + x_4 = 200 + 0 - 106.1 - 259.8 = -165.9 \, \text{m}
  \]

- **Total y-component:**
  \[
  R_y = y_1 + y_2 + y_3 + y_4 = 0 + 100 + 106.1 - 150 = 56.1 \, \text{m}
  \]

#### Step 3: Calculate the magnitude of the resultant vector
The magnitude of the displacement is given by:
\[
\text{Displacement} = \sqrt{R_x^2 + R_y^2}
\]
\[
\text{Displacement} = \sqrt{(-165.9)^2 + (56.1)^2} \approx \sqrt{27529.61 + 3147.21} \approx \sqrt{30676.82} \approx 175.1 \, \text{m}
\]

### Final Answer:
The magnitude of the total displacement is approximately **175.1 m**.

Would you like a step-by-step drawing guide or have any other questions? Here are some related questions you might find interesting:

1. How do you calculate the direction of the resultant displacement vector?
2. What is the difference between vector addition graphically and algebraically?
3. How do you resolve a vector into its perpendicular components?
4. How do you find the resultant vector when given in polar coordinates?
5. What is the significance of displacement in physics versus distance?

**Tip:** When adding vectors, always resolve them into their x and y components before summing them to avoid mistakes.

Learn how to calculate the magnitude of total displacement using vector addition and components. This problem involves multiple displacement vectors: 200m East, 100m North, 150m Northwest, and 300m South 30 degrees West. Detailed step-by-step solution and formulas included. Suitable for Grades 11-12.

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