What is the passenger speed wre ground a) An ocean liner is travelling $$ 18 \mathrm{~km} / \mathrm{h} $$ due south. A passenger on the deck walks toward the rear of the ship at $$ 3.0 \mathrm{~m} / \mathrm{s} $$. b) The same ship as in \#2, but the passenger has turned right and is walking towards the railing. b.

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**Part (a):**

1. Convert the ship’s speed from kilometers per hour to meters per second. Since  
   $$
   1\,\text{km/h} \approx 0.2778\,\text{m/s},
   $$  
   we have  
   $$
   18\,\text{km/h} \approx 18 \times 0.2778\,\text{m/s} = 5\,\text{m/s}.
   $$

2. The ship is travelling due south at $$5\,\text{m/s}$$. Assuming the ship’s forward (bow) is in the direction of travel, walking toward the rear means moving in the opposite direction, i.e. due north, at $$3.0\,\text{m/s}$$.

3. The passenger’s velocity relative to the ground is the vector sum of the ship’s velocity and the passenger’s velocity relative to the ship:
   $$
   \vec{v}_{\text{passenger}} = \vec{v}_{\text{ship}} + \vec{v}_{\text{relative}}.
   $$
   Here:
   - $$\vec{v}_{\text{ship}} = 5\,\text{m/s}$$ (south), and  
   - $$\vec{v}_{\text{relative}} = 3.0\,\text{m/s}$$ (north).

4. Since north and south are opposite directions, subtract the speeds:
   $$
   v_{\text{passenger}} = 5\,\text{m/s} - 3.0\,\text{m/s} = 2\,\text{m/s},
   $$
   and the direction is south.

Thus, the passenger’s speed relative to the ground is  
$$
2\,\text{m/s}\ \text{south}.
$$

---

**Part (b):**

1. The ship’s speed remains $$5\,\text{m/s}$$ due south.

2. In this scenario, the passenger turns right relative to the ship’s forward (south) direction. For a ship travelling south, “right” corresponds to the westward direction. The passenger walks toward the railing at a speed of $$3.0\,\text{m/s}$$ relative to the ship.

3. Now, the passenger’s velocity relative to the ground is given by the vector sum:
   $$
   \vec{v}_{\text{passenger}} = \vec{v}_{\text{ship}} + \vec{v}_{\text{relative}},
   $$
   where
   $$
   \vec{v}_{\text{ship}} = 5\,\text{m/s}\ \text{(south)} \quad \text{and} \quad \vec{v}_{\text{relative}} = 3.0\,\text{m/s}\ \text{(west)}.
   $$

4. Set up a coordinate system with south as the negative $$y$$-direction and west as the negative $$x$$-direction. Then, the components are:
   - $$v_{y} = -5\,\text{m/s}$$,
   - $$v_{x} = -3.0\,\text{m/s}$$.

5. The magnitude of the resultant velocity is:
   $$
   v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.0)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83\,\text{m/s}.
   $$

6. The direction relative to due south (the $$y$$-axis) can be found by:
   $$
   \theta = \tan^{-1}\!\left(\frac{3.0}{5}\right) \approx \tan^{-1}(0.6) \approx 31^\circ.
   $$
   This angle is measured toward the west.

Thus, the passenger’s speed relative to the ground in part (b) is approximately  
$$
5.83\,\text{m/s} \quad \text{at}\ 31^\circ\ \text{west of south}.
$$
**Part (a):** The passenger is moving at 2 m/s south relative to the ground. **Part (b):** The passenger is moving at approximately 5.83 m/s, 31 degrees west of south, relative to the ground.

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