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.

**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 \[ 5.83\,\text{m/s} \quad \text{at}\ 31^\circ\ \text{west of south}. \]