Step 1: Set up coordinate system and define vectors. Let the initial point be the origin. Let East be the positive x-axis and North be the positive y-axis. The first displacement vector is 4 km towards West: d⃗₁ = -4î km.

Step 2: Resolve the second displacement. The second displacement is 3 km, 30° east of north. This means the angle with the positive y-axis is 30° towards the positive x-axis. The angle with the positive x-axis is 90° - 30° = 60°. The components are: x-component = 3 cos 60° = 3 × ½ = 1.5 km; y-component = 3 sin 60° = 3 × ($\sqrt{3}$/2) = 3$\sqrt{3}$/2 km. So, d⃗₂ = 1.5î + (3$\sqrt{3}$/2)ĵ km.

Step 3: Find the resultant displacement vector. Using the triangle law of vector addition: D⃗ = d⃗₁ + d⃗₂ = (-4î) + (1.5î + 3$\sqrt{3}$/2 ĵ) = (-4 + 1.5)î + (3$\sqrt{3}$/2)ĵ = -2.5î + (3$\sqrt{3}$/2)ĵ km.

Step 4: Calculate the magnitude of displacement. |D⃗| = √[(-2.5)² + (3$\sqrt{3}$/2)²] = √[6.25 + 27/4] = √[6.25 + 6.75] = $\sqrt{13}$ km $\approx$ 3.6 km.

Step 5: Determine the direction. The resultant vector has a negative x-component (west) and positive y-component (north). The angle θ with the West direction is given by: tan θ = |y-component| / |x-component| = (3$\sqrt{3}$/2) / 2.5 = 3$\sqrt{3}$/5. Therefore, θ = tan⁻¹(3$\sqrt{3}$/5) $\approx$ 46°. The displacement is $\sqrt{13}$ km at an angle of tan⁻¹(3$\sqrt{3}$/5) north of west.