Let's establish a coordinate system where the positive x-direction is East and the positive y-direction is North. The velocity vectors of the boats are:

$\vec{v}_A = v_A \hat{i}$ (Boat A travelling East) $\vec{v}_B = v_B \hat{j}$ (Boat B travelling North)

The velocity of Boat B relative to Boat A, $\vec{v}_{B/A}$, is found by the vector subtraction:

$$\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A$$

Substituting the vector components:

$$\vec{v}_{B/A} = v_B \hat{j} - v_A \hat{i} = -v_A \hat{i} + v_B \hat{j}$$

[Q1] Find the speed of Boat B as observed by a person on Boat A. The speed is the magnitude of the relative velocity vector, $v_{B/A} = |\vec{v}_{B/A}|$. Using the Pythagorean theorem:

$$v_{B/A} = \sqrt{(-v_A)^2 + (v_B)^2} = \sqrt{v_A^2 + v_B^2}$$

Substituting the given values, $v_A = 30$ km/h and $v_B = 45$ km/h:

$$v_{B/A} = \sqrt{(30)^2 + (45)^2} = \sqrt{900 + 2025} = \sqrt{2925}$$ $$v_{B/A} \approx 54.1 \text{ km/h}$$

[Q2] Find the direction of Boat B as observed by a person on Boat A. The direction is given by the angle $\theta$ of the vector $\vec{v}_{B/A} = -v_A \hat{i} + v_B \hat{j}$. The vector points West (negative x-direction) and North (positive y-direction). Let's find the angle $\theta$ North of West.

$$\tan(\theta) = \frac{\text{North component}}{\text{West component}} = \frac{v_B}{v_A}$$

Substituting the values:

$$\tan(\theta) = \frac{45}{30} = 1.5$$ $$\theta = \arctan(1.5)$$ $$\theta \approx 56.3^\circ$$

The direction is $56.3^\circ$ North of West.