P is a point on the southern bank of a river - Leaving Cert Applied Maths - Question 2 - 2019

For ship B, the speed is 58 km/h and the angle eta is defined by $an \beta = \frac{20}{21}$.

Using similar trigonometric calculations:

• North component: $58 \times \frac{20}{\text{hypotenuse}}$
• East component: $58 \times \frac{21}{\text{hypotenuse}}$

Thus, the velocity of ship B is: $\mathbf{V_B} = 58 \sin(\beta) \mathbf{i} + 58 \cos(\beta) \mathbf{j} = 40 \mathbf{i} - 42 \mathbf{j}$

The relative velocity of ship A to ship B is given by: $\mathbf{V_{AB}} = \mathbf{V_A} - \mathbf{V_B}$

Substituting the previously calculated values: $\mathbf{V_{AB}} = (60 \mathbf{i} + 32 \mathbf{j}) - (40 \mathbf{i} - 42 \mathbf{j})$ This simplifies to: $\mathbf{V_{AB}} = 20 \mathbf{i} + 74 \mathbf{j}$

To find the time taken for ship A to reach point R, which is 30 km downstream from P, we use: $t = \frac{\text{Distance}}{\text{Speed}} = \frac{30}{60} = 0.5 \text{ hours}$

If ship A takes 0.5 hours, ship B takes: $\text{Time for B} = 0.5 + \frac{12}{60} = 0.5 + 0.2 = 0.7 \text{ hours}$

Thus, using the speed of ship B: $d = 30 + 40(0.5) = 2 \text{ km}$

Using the previous computations, the width of the river is calculated as: $w = 32(0.5) + 42(0.7) = 45.4 \text{ km}$