We define our coordinate system such that the positive X-axis points East and the positive Y-axis points North.

Here's a step-by-step solution to the problem:

We define our coordinate system such that the positive X-axis points East and the positive Y-axis points North.

Part (i): The components of each displacement

• Displacement 1 ($\vec{d_1}$): 4.0 m South West South West means the angle is $225^\circ$ from the positive X-axis, or $45^\circ$ South of West.

  $$d_{1x} = 4.0 \cos(225^\circ) = 4.0 \left(-\frac{\sqrt{2}}{2}\right) = -2\sqrt{2} \text{ m} \approx -2.83 \text{ m}$$ $$d_{1y} = 4.0 \sin(225^\circ) = 4.0 \left(-\frac{\sqrt{2}}{2}\right) = -2\sqrt{2} \text{ m} \approx -2.83 \text{ m}$$

  So, $\vec{d_1} = \boxed{(-2\sqrt{2} \text{ m}, -2\sqrt{2} \text{ m})}$

• Displacement 2 ($\vec{d_2}$): 5.0 m East This displacement is entirely along the positive X-axis.

  $$d_{2x} = 5.0 \text{ m}$$ $$d_{2y} = 0 \text{ m}$$

  So, $\vec{d_2} = \boxed{(5.0 \text{ m}, 0 \text{ m})}$

• Displacement 3 ($\vec{d_3}$): 6.0 m in a direction 30° E of N "30° E of N" means $30^\circ$ from the North axis towards the East. This corresponds to an angle of $90^\circ - 30^\circ = 60^\circ$ from the positive X-axis.

  $$d_{3x} = 6.0 \cos(60^\circ) = 6.0 \left(\frac{1}{2}\right) = 3.0 \text{ m}$$ $$d_{3y} = 6.0 \sin(60^\circ) = 6.0 \left(\frac{\sqrt{3}}{2}\right) = 3\sqrt{3} \text{ m} \approx 5.20 \text{ m}$$

  So, $\vec{d_3} = \boxed{(3.0 \text{ m}, 3\sqrt{3} \text{ m})}$

Part (ii): The components of the resultant displacement

Let the resultant displacement be $\vec{R}$. We sum the respective components:

$$R_x = d_{1x} + d_{2x} + d_{3x}$$ $$R_x = -2\sqrt{2} + 5.0 + 3.0 = (8.0 - 2\sqrt{2}) \text{ m}$$ $$R_y = d_{1y} + d_{2y} + d_{3y}$$ $$R_y = -2\sqrt{2} + 0 + 3\sqrt{3} = (3\sqrt{3} - 2\sqrt{2}) \text{ m}$$

Numerically:

$$R_x \approx 8.0 - 2(1.414) = 8.0 - 2.828 = 5.172 \text{ m}$$ $$R_y \approx 3(1.732) - 2(1.414) = 5.196 - 2.828 = 2.368 \text{ m}$$

The components of the resultant displacement are $\boxed{(5.17 \text{ m}, 2.37 \text{ m})}$.

Part (iii): The magnitude and direction of the resultant displacement

The magnitude of the resultant displacement $|\vec{R}|$ is:

$$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$$ $$|\vec{R}| = \sqrt{(5.172)^2 + (2.368)^2}$$ $$|\vec{R}| = \sqrt{26.75 + 5.61} = \sqrt{32.36} \approx 5.69 \text{ m}$$

The direction $\theta$ is given by:

$$\theta = \arctan\left(\frac{R_y}{R_x}\right)$$ $$\theta = \arctan\left(\frac{2.368}{5.172}\right) = \arctan(0.4578) \approx 24.6^\circ$$

Since both $R_x$ and $R_y$ are positive, the resultant displacement is in the first quadrant.
The magnitude of the resultant displacement is $\boxed{5.69 \text{ m}}$ and its direction is $\boxed{24.6^\circ \text{ North of East}}$.

Part (iv): The displacement that would be required to bring the rat back to the starting point

To bring the rat back to the starting point, the required displacement must be equal in magnitude and opposite in direction to the resultant displacement. Let this be $\vec{D}_{return}$.

$$\vec{D}_{return} = -\vec{R}$$ $$D_{return, x} = -R_x = -(8.0 - 2\sqrt{2}) \text{ m} \approx -5.17 \text{ m}$$ $$D_{return, y} = -R_y = -(3\sqrt{3} - 2\sqrt{2}) \text{ m} \approx -2.37 \text{ m}$$

The magnitude of this displacement is the same as $|\vec{R}|$, which is approximately $5.69 \text{ m}$.

To find the direction, we note that both components are negative, placing the displacement in the third quadrant (South-West). The angle $\alpha$ relative to the negative X-axis (West) is:

$$\alpha = \arctan\left(\frac{|D_{return, y}|}{|D_{return, x}|}\right) = \arctan\left(\frac{2.368}{5.172}\right) \approx 24.6^\circ$$

The displacement required to bring the rat back to the starting point has a magnitude of $\boxed{5.69