1.3 Two ships leave Cape Town harbour simultaneously. Ship K sails North West at \( 50 \mathrm{~km} / \mathrm{h} \) and ship L sails South West at \( 70 \mathrm{~km} / \mathrm{h} \). Calculate the velocity of ship L relative to the velocity of ship K magnitude and direction.

To find the velocity of ship L relative to ship K, we first need to determine the velocity vectors of both ships. For Ship K, which sails North West at 50 km/h: - Its velocity vector can be represented as: \[ \mathbf{V_K} = ( -\frac{50}{\sqrt{2}}, \frac{50}{\sqrt{2}} ) \approx (-35.36, 35.36) \, \text{km/h} \] For Ship L, which sails South West at 70 km/h: - Its velocity vector can be represented as: \[ \mathbf{V_L} = ( -\frac{70}{\sqrt{2}}, -\frac{70}{\sqrt{2}} ) \approx (-49.50, -49.50) \, \text{km/h} \] Now we can find the relative velocity of ship L with respect to ship K by subtracting Ship K's velocity from Ship L's velocity: \[ \mathbf{V_{L/K}} = \mathbf{V_L} - \mathbf{V_K} = \left(-49.50 + 35.36, -49.50 - 35.36\right) \approx (-14.14, -84.86) \, \text{km/h} \] Next, we calculate the magnitude of this velocity vector: \[ |\mathbf{V_{L/K}}| = \sqrt{(-14.14)^2 + (-84.86)^2} \approx \sqrt{200.02 + 7202.40} \approx \sqrt{7402.42} \approx 86.03 \, \text{km/h} \] For the direction, we calculate the angle using the tangent function: \[ \tan(\theta) = \frac{-84.86}{-14.14} \rightarrow \theta \approx \tan^{-1}(6) \approx 80.54^{\circ} \] Since both components are negative, the direction is in the third quadrant: \[ \text{Direction: } \theta = 180^{\circ} + 80.54^{\circ} \approx 260.54^{\circ} \text{ from North.} \] So, the velocity of ship L relative to ship K has a magnitude of approximately \( 86.03 \, \text{km/h} \) and a direction of \( 260.54^{\circ} \).