Rain is falling with a speed of 25 m s⁻¹ at an angle of 20° to the vertical - Leaving Cert Applied Maths - Question 2 - 2012

Given:

• Ship A: 50 km north of ship B.
• Ship A velocity: 24/√2 km h⁻¹ southwest.
• Ship B velocity: 17 km h⁻¹ due west.

1. Define the velocity vectors:

   • Velocity of Ship A:

   $\vec{V_A} = (17 \hat{i}) + ( -24 \hat{j})$

   • Velocity of Ship B: $\vec{V_B} = ( -24/\sqrt{2} \hat{i}) + 0 \hat{j}$

2. Find relative velocity:

   $\vec{V_{B/A}} = \vec{V_B} - \vec{V_A} = (17 \hat{i} - (-24/\sqrt{2}) \hat{i}) + ( -24 \hat{j}) = (7 \hat{i} + 24 - 17) \hat{j} = 7 \hat{i} + 7 \hat{j}$

3. Calculate magnitude:

   $|\vec{V_{B/A}}| = \sqrt{7^2 + 24^2} = 25 ext{ km h}^{-1}$

4. Calculate the direction:

   $\theta = \tan^{-1}(\frac{7}{24})$

   • The direction of the velocity of B relative to A is approximately: [ 16.26° \text{ from the west towards north} ]

1. Determine distance between ships:

   • At noon A is 50 km north of B.

2. Calculate time to be within 20 km:

   • Distance between them tends to reduce, and they can exchange signals when they are no more than 20 km apart.

3. Calculation part of relative speed and time taken:

   • [ |d_C| = |AB| - |BC| ]

   • Total covered distance: 50 km - 20 km = 30 km.

   • From relative speed found earlier, find the time is given by: $t = \frac{30 \text{ km}} {25 \text{ km h}^{-1}} = 1.2 ext{ hours}$

   • Add that to current time of 12:00:

   • Therefore they can exchange signals at approximately 1:20 PM.

1. Initial distance:

   • Ship A is initially 50 km north of B, and they can maintain communication until they are more than 20 km apart.

2. Calculate distance reduction:

   • When the ships reach a distance of 20 km apart, relative speed is constant.

3. Calculate the time until they go beyond distance:

   • The rate at which distance reduces is determined by:
   $$- Thus, by calculations, they can continue exchanging signals for approximately **1h 9min** after their first exchange.$$