A river is 100 metres wide and has parallel banks - Leaving Cert Applied Maths - Question 2 - 2010

The velocity of B relative to C is calculated as: $\mathbf{v}_{BC} = \mathbf{v}_B - \mathbf{v}_C$ Substituting the values: $\mathbf{v}_{BC} = (5\mathbf{i} + 12\mathbf{j}) - (0\mathbf{i} + 3\mathbf{j}) = 5\mathbf{i} + 9\mathbf{j}$

To find the magnitude: $|\mathbf{v}_{BC}| = \sqrt{5^2 + 9^2} = \sqrt{25 + 81} = \sqrt{106} \approx 10.3 \text{ m/s}$

To find the direction, we can use: $\tan(\theta) = \frac{9}{5}$ Thus, $\theta = \tan^{-1}\left(\frac{9}{5}\right) \approx 60.9^\circ \text{ N of E}.$

The time it takes for B to cross the river is given by: $\text{time} = \frac{\text{distance}}{\text{velocity}} = \frac{100 \text{ m}}{5 \text{ m/s}} = 20 \text{ s}.$

To calculate the distance |𝑃𝑄|, we use: $|\mathbf{P}Q| = 2(0i + 3\text{ m}) + \sqrt{5^2 + 9^2} = 13 \text{ m/s}.$ Hence, the total distance from P to Q is: $|PQ| = 20 \text{ s} \cdot 13 \text{ m/s} = 260 \text{ m}.$