We establish a coordinate system with the +x-axis pointing East and the +y-axis pointing North. The port is at the origin. The velocities of the ships are represented as vectors.

Ship A (Northwest at 24 knots): The direction Northwest corresponds to an angle of $\theta_A = 135°$ with the positive x-axis.

$$ \vec{v}_A = v_A (\cos\theta_A \hat{i} + \sin\theta_A \hat{j}) = 24 (\cos 135° \hat{i} + \sin 135° \hat{j}) $$ $$ \vec{v}_A = 24(-\frac{\sqrt{2}}{2}\hat{i} + \frac{\sqrt{2}}{2}\hat{j}) = (-12\sqrt{2} \hat{i} + 12\sqrt{2} \hat{j}) \text{ knots} $$

Ship B (40° West of South at 28 knots): This direction corresponds to an angle of $\theta_B = 270° - 40° = 230°$ with the positive x-axis.

$$ \vec{v}_B = v_B (\cos\theta_B \hat{i} + \sin\theta_B \hat{j}) = 28 (\cos 230° \hat{i} + \sin 230° \hat{j}) \text{ knots} $$

[Q1] & [Q2] Velocity of Ship A Relative to B

The velocity of ship A relative to ship B, $\vec{v}_{A/B}$, is given by the vector difference:

$$ \vec{v}_{A/B} = \vec{v}_A - \vec{v}_B $$

The components of the relative velocity are:

$$ v_{A/B,x} = v_{Ax} - v_{Bx} = -12\sqrt{2} - 28\cos 230° $$ $$ v_{A/B,y} = v_{Ay} - v_{By} = 12\sqrt{2} - 28\sin 230° $$

Substituting the numerical values:

$$ v_{A/B,x} \approx -16.97 - 28(-0.643) = -16.97 + 18.00 = 1.03 \text{ knots} $$ $$ v_{A/B,y} \approx 16.97 - 28(-0.766) = 16.97 + 21.45 = 38.42 \text{ knots} $$

So, $\vec{v}_{A/B} \approx (1.03\hat{i} + 38.42\hat{j})$ knots.

[Q1] The magnitude of the relative velocity is:

$$ |\vec{v}_{A/B}| = \sqrt{v_{A/B,x}^2 + v_{A/B,y}^2} $$ $$ |\vec{v}_{A/B}| \approx \sqrt{(1.03)^2 + (38.42)^2} \approx 38.4 \text{ knots} $$

[Q2] The direction of the relative velocity is given by the angle $\theta_{A/B}$:

$$ \theta_{A/B} = \arctan\left(\frac{v_{A/B,y}}{v_{A/B,x}}\right) $$ $$ \theta_{A/B} \approx \arctan\left(\frac{38.42}{1.03}\right) \approx 88.5° $$

This angle is measured counter-clockwise from the East (+x axis). To express this as a bearing from North, we calculate $90° - 88.5° = 1.5°$. Since both components are positive, the direction is East of North. The direction is N 1.5° E.

[Q3] Time until Ships are 160 Nautical Miles Apart

The separation vector between the ships is $\vec{r}_{A/B} = \vec{r}_A - \vec{r}_B$. Since they start from the same point, $\vec{r}_A = \vec{v}_A t$ and $\vec{r}_B = \vec{v}_B t$.

$$ \vec{r}_{A/B} = (\vec{v}_A - \vec{v}_B)t = \vec{v}_{A/B} t $$

The separation distance $d$ is the magnitude of this vector:

$$ d = |\vec{r}_{A/B}| = |\vec{v}_{A/B}| t $$

Solving for time $t$:

$$ t = \frac{d}{|\vec{v}_{A/B}|} $$

Substituting the given values, $d = 160$ nautical miles and $|\vec{v}_{A/B}| \approx 38.4$ knots:

$$ t \approx \frac{160}{38.4} \approx 4.16 \text{ hours} $$

[Q4] Bearing of B Relative to A at That Time

The position of B relative to A is given by the vector $\vec{r}_{B/A} = \vec{r}_B - \vec{r}_A = -\vec{r}_{A/B} = -\vec{v}_{A/B} t$. The direction of this vector is constant and is opposite to the direction of $\vec{v}_{A/B}$. The direction of $\vec{v}_{A/B}$ was N 1.5° E. The opposite direction is S 1.5° W. As a three-figure bearing measured clockwise from North, this corresponds to an angle of $180° + 1.5° = 181.5°$.