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.