A river is 57 metres wide and has parallel banks - Leaving Cert Applied Maths - Question 2 - 2013

The velocity of B relative to C is calculated as follows:

$\mathbf{v}_{bc} = \mathbf{v}_b - \mathbf{v}_c$

Substituting in the known values:

$\mathbf{v}_{bc} = (4 \boldsymbol{i} - 3 \boldsymbol{j}) - (7 \boldsymbol{i} + 0 \boldsymbol{j})$

This simplifies to:

$\mathbf{v}_{bc} = (4 - 7) \boldsymbol{i} + (-3) \boldsymbol{j} = -3 \boldsymbol{i} - 3 \boldsymbol{j}$

To find the magnitude of $\mathbf{v}_{bc}$, we use the formula:

$||\mathbf{v}_{bc}|| = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}$

For the direction, we calculate:

$\theta = \tan^{-1}\left(\frac{-3}{-3}\right) = \tan^{-1}(1) = 45^{\circ}$

Since both components are negative, the angle lies in the third quadrant, giving a direction of:

$\text{S} 45^{\circ} \text{ W}$

The width of the river is given as 57 meters. To find the time taken to cross, we can use the following relationship:

$t = \frac{d}{v}$

Where $d = 57 ext{ m}$ and $v = 3 ext{ m/s}$ (the vertical component of velocity B relative to C). Thus:

$t = \frac{57}{3} = 19 ext{ seconds}$

The distance $|PQ|$ can be calculated using:

$|PQ| = v_{bc} \cdot t = 5 \sqrt{2} \cdot 19$

Approximately, this gives:

$|PQ| = 95 ext{ m}$