A torpedo is moving at 32 m/s NW. A freighter is moving at 10 m/s, 60 degrees East of South. Calculate the relative velocity of the torpedo with respect to the freighter.

Step 1: Define the velocities in component form.
Let the positive x-axis be East and the positive y-axis be North.

The velocity of the torpedo, $\vec{v}_T$, is $32 \text{ m/s NW}$.
NW means $45^\circ$ North of West. This corresponds to an angle of $135^\circ$ from the positive x-axis.

$$\vec{v}_T = (32 \cos 135^\circ) \hat{i} + (32 \sin 135^\circ) \hat{j}$$ $$\vec{v}_T = \left(32 \times -\frac{\sqrt{2}}{2}\right) \hat{i} + \left(32 \times \frac{\sqrt{2}}{2}\right) \hat{j}$$ $$\vec{v}_T = -16\sqrt{2} \hat{i} + 16\sqrt{2} \hat{j} \text{ m/s}$$

Numerically:

$$v_{Tx} \approx -16 \times 1.4142 = -22.627 \text{ m/s}$$ $$v_{Ty} \approx 16 \times 1.4142 = 22.627 \text{ m/s}$$

The velocity of the freighter, $\vec{v}_F$, is $10 \text{ m/s } 60^\circ \text{ E of S}$.
This means $60^\circ$ East from the South direction. South is $-90^\circ$ or $270^\circ$. So the angle from the positive x-axis is $270^\circ + 60^\circ = 330^\circ$, or equivalently $-30^\circ$.

$$\vec{v}_F = (10 \cos (-30^\circ)) \hat{i} + (10 \sin (-30^\circ)) \hat{j}$$ $$\vec{v}_F = \left(10 \times \frac{\sqrt{3}}{2}\right) \hat{i} + \left(10 \times -\frac{1}{2}\right) \hat{j}$$ $$\vec{v}_F = 5\sqrt{3} \hat{i} - 5 \hat{j} \text{ m/s}$$

Numerically:

$$v_{Fx} \approx 5 \times 1.7321 = 8.660 \text{ m/s}$$ $$v_{Fy} = -5 \text{ m/s}$$

Step 2: Calculate the relative velocity of the torpedo with respect to the freighter, $\vec{v}_{T/F}$.
The relative velocity is given by $\vec{v}_{T/F} = \vec{v}_T - \vec{v}_F$.

$$\vec{v}_{T/F} = (-16\sqrt{2} \hat{i} + 16\sqrt{2} \hat{j}) - (5\sqrt{3} \hat{i} - 5 \hat{j})$$ $$$$