Let's break down the problem step-by-step: --- ## **Step 1: Assign Velocities** - **Velocity of Submarine, \( \vec{V}_S \):** - Magnitude: \( 30 \) km/h - Direction: North \( 30^\circ \) East - **Velocity of Torpedo relative to ground, \( \vec{V}_T \):** - Magnitude: \( 80 \) km/h - Direction: Due North We are to find the **velocity of the torpedo relative to the submarine**, \( \vec{V}_{T/S} \). --- ## **Step 2: Write Relative Velocity Equation** \[ \vec{V}_{T/S} = \vec{V}_T - \vec{V}_S \] --- ## **Step 3: Resolve Velocities into Components** Let's use: - \( +y \): North - \( +x \): East ### **Submarine (\(\vec{V}_S\))** - North \( 30^\circ \) East means \( 30^\circ \) east of north. \[ V_{S,y} = 30 \cos 30^\circ = 30 \times \frac{\sqrt{3}}{2} = 25.98 \text{ km/h} \] \[ V_{S,x} = 30 \sin 30^\circ = 30 \times \frac{1}{2} = 15 \text{ km/h} \] ### **Torpedo (\(\vec{V}_T\))** - Only North component (since fired due north): \[ V_{T,y} = 80 \text{ km/h} \] \[ V_{T,x} = \text{ km/h} \] --- ## **Step 4: Find Relative Velocity Components** \[ V_{T/S, y} = V_{T,y} - V_{S,y} = 80 - 25.98 = 54.02 \text{ km/h} \] \[ V_{T/S, x} = V_{T,x} - V_{S,x} = - 15 = -15 \text{ km/h} \] --- ## **Step 5: Find Magnitude of Relative Velocity** \[ |\vec{V}_{T/S}| = \sqrt{(V_{T/S, x})^2 + (V_{T/S, y})^2} \] \[ = \sqrt{(-15)^2 + (54.02)^2} \] \[ = \sqrt{225 + 2918.16} \] \[ = \sqrt{3143.16} \] \[ = 56.1 \text{ km/h} \] --- ## **Step 6: Find Direction of Relative Velocity** \[ \tan \theta = \frac{|V_{T/S, x}|}{V_{T/S, y}} = \frac{15}{54.02} \] \[ \theta = \arctan \left(\frac{15}{54.02}\right) \] \[ \theta = \arctan (.2776) \] \[ \theta \approx 15.5^\circ \] Since \(V_{T/S, x}\) is negative, the direction is **west of north**. --- ## **Step 7: Final Answer** ### **Magnitude:** \[ \boxed{56.1 \text{ km/h}} \] ### **Direction:** \[ \boxed{15.5^\circ \text{ west of north}} \] --- ### **Summary Table** | Quantity | Value | |---|---| | Magnitude of relative velocity | \( 56.1 \) km/h | | Direction | \( 15.5^\circ \) west of north | --- **Therefore, the velocity of the torpedo relative to the submarine is \( \boxed{56.1\,\text{km/h}, \ 15.5^\circ\,\text{west of north}} \).**