Two cars, A and B, are located 100 m and 40 m respectively from junction O, as shown in the diagram - Leaving Cert Applied Maths - Question 2 - 2021

The velocity of car A, traveling east, is represented as:

$\mathbf{V}_A = 5\mathbf{i} + 0\mathbf{j}$

The velocity of car B, traveling north, is represented as:

$\mathbf{V}_B = 0\mathbf{i} + 8\mathbf{j}$

To find the relative velocity of A with respect to B, we use the following equation:

$\mathbf{V}_{AB} = \mathbf{V}_A - \mathbf{V}_B$

Substituting the velocities:

$\mathbf{V}_{AB} = (5\mathbf{i} + 0\mathbf{j}) - (0\mathbf{i} + 8\mathbf{j}) = 5\mathbf{i} - 8\mathbf{j}$

To find the magnitude of this velocity:

$|\mathbf{V}_{AB}| = \sqrt{(5)^2 + (-8)^2} = \sqrt{25 + 64} = \sqrt{89}$

For the direction, we can find the angle using:

$\tan(\theta) = \frac{-8}{5}$

Therefore, the angle is:

$\theta = \tan^{-1}\left(\frac{-8}{5}\right) = 58°\text{ (approximately)}$

To determine the heading direction of the aircraft, we need to adjust for the wind speed. The ground velocity needed for the aircraft to reach D is calculated as follows:

Let:

• $v = 400 km / 17 h = 235.29 km/h$

The angle can be calculated using trigonometry:

$\sin(\alpha) = \frac{50}{235.29}$

Thus:

$\alpha = \sin^{-1}\left(\frac{50}{235.29}\right) \approx 12.27°$

To find the time of landing, we first calculate the distance traveled:

$x = \frac{400}{229.92} \approx 1.74 h$

Since the aircraft leaves at 13:00, adding approximately 1.74 hours results in:

$\text{Time of landing} = 14:44$