(a) A girl cycles at a constant speed of 5 m s^-1 on level ground - Leaving Cert Applied Maths - Question 2 - 2017

To find the velocity of ship Q relative to ship P, we consider the velocity components:

1. Ship P's velocity is: $\bar{v}_{P} = 15\hat{j} \, km/h$

2. The velocity of ship Q can then be assessed using its directed motion (east 60° south at 15√3 km/h):

   $\bar{v}_{Q} = 15\hat{i} \cos(60) - 15\sqrt{3}\hat{j} \sin(60) + 15\hat{j}$

   Simplifying this gives:

   • Velocity component towards east: $= \frac{15√3}{2} \hat{i} - \frac{15\sqrt{3}}{2} \hat{j} + 15\hat{j}$

   Therefore,

   • The total velocity of Q: $\bar{v}_{Q} = \left(\frac{15\sqrt{3}}{2} \hat{i} + \left(15 - \frac{5}{2}\right)\hat{j}\right) = 15 km/h, E30° S.$

To find the angle $\theta$ after ship P and Q are closest:

1. We establish the positions of ships using relative motion:
   • Let $R$ represent the closest distance, with the placement of Q from P being: $|\bar{R}| = 10\sin(60) = 5√3.$
2. After evaluating, we find: $\tan(\alpha) = \frac{5\sqrt{3}}{10}$
   • Thus, $\alpha = 84.3°.$
3. Given that the total angle concerning the fixed direction is: $\theta = 180° - 84.3° - 60° = 35.7°.$

The calculation of distance after 12 minutes involves:

1. The relative distance over time from the position of ship Q:

   • Calculating for x: $x = 15v \left(\frac{12}{60}\right)\ = 1.5 km$

2. Combine with the position:

   • Total distance at the end of the assessment period: $d = \sqrt{(1.5)^{2} + (\frac{5}{3})^{2}} \approx 9.04 km.$