A ship `A` streams due north at `16 km h^-1` and a ship `B` due west at `12 km h^-1`. At a certain instant `B` is `10 km` north east of `A`. Find the
(a) magnitude of velocity of `A relative to B`.
(b) nearest distance of approach of ships.

Explanation

To solve the problem, we need to find the relative velocity of ship A with respect to ship B and the nearest distance of approach of the ships.

### Step-by-step Solution:

#### Given Data:
- Velocity of ship A, \( V_A = 16 \, \text{km/h} \) (due north)
- Velocity of ship B, \( V_B = 12 \, \text{km/h} \) (due west)
- Initial position of ship B is \( 10 \, \text{km} \) northeast of ship A.

#### (a) Magnitude of Velocity of A Relative to B

1. **Determine the velocity vectors:**
- Ship A's velocity vector can be represented as:
\[ \vec{V_A} = 0 \hat{i} + 16 \hat{j} \, \text{km/h} \]
- Ship B's velocity vector can be represented as:
\[ \vec{V_B} = -12 \hat{i} + 0 \hat{j} \, \text{km/h} \]

2. **Calculate the relative velocity of A with respect to B:**
\[ \vec{V_{AB}} = \vec{V_A} - \vec{V_B} = (0 \hat{i} + 16 \hat{j}) - (-12 \hat{i} + 0 \hat{j}) = 12 \hat{i} + 16 \hat{j} \, \text{km/h} \]

3. **Find the magnitude of the relative velocity:**
\[ |\vec{V_{AB}}| = \sqrt{(12)^2 + (16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, \text{km/h} \]

#### (b) Nearest Distance of Approach of Ships

1. **Determine the angle of the relative velocity:**
- The angle \( \alpha \) can be calculated using:
\[ \tan \alpha = \frac{16}{12} = \frac{4}{3} \]
- Therefore, \( \alpha = \tan^{-1} \left( \frac{4}{3} \right) \).

2. **Calculate the distance of closest approach:**
- The initial distance \( d \) between the ships is \( 10 \, \text{km} \) at an angle of \( 45^\circ \) (northeast).
- The nearest distance \( D \) can be calculated using the formula:
\[ D = d \cdot \sin \alpha \]
- Where \( d = 10 \, \text{km} \) and \( \sin \alpha = \frac{4}{5} \) (from the triangle formed by the velocities).

3. **Substituting the values:**
\[ D = 10 \cdot \frac{4}{5} = 8 \, \text{km} \]

4. **Calculate the perpendicular distance:**
- The perpendicular distance from the path of A to the initial position of B can be calculated using:
\[ D_{\text{min}} = 10 \cdot \cos(45^\circ) - 10 \cdot \sin(45^\circ) \cdot \frac{12}{20} \]
- Simplifying gives us:
\[ D_{\text{min}} = 10 \cdot \frac{1}{\sqrt{2}} - 10 \cdot \frac{4}{5} \cdot \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} - \frac{8}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \, \text{km} \]

### Final Answers:
- (a) The magnitude of the velocity of A relative to B is \( 20 \, \text{km/h} \).
- (b) The nearest distance of approach of the ships is \( 2 \, \text{km} \).