2. 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 \( 3.0 \mathrm{~m} / \mathrm{s} \).

To solve this problem, we need to determine the velocity of the passenger relative to the ground. We have two velocities to consider: 1. The velocity of the ocean liner, which is traveling due south at \( 18 \, \text{km/h} \). 2. The velocity of the passenger walking toward the rear of the ship at \( 3.0 \, \text{m/s} \). ### Step 1: Convert the speed of the ocean liner to the same units as the passenger's speed. The speed of the ocean liner is given in kilometers per hour (km/h), and we need to convert it to meters per second (m/s) for consistency. The conversion factor is: \[ 1 \, \text{km/h} = \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{1}{3.6} \, \text{m/s} \] Now, we can convert \( 18 \, \text{km/h} \) to m/s: \[ v_{\text{liner}} = 18 \, \text{km/h} \times \frac{1}{3.6} \, \text{m/s} \approx 5 \, \text{m/s} \] ### Step 2: Determine the direction of the velocities. - The ocean liner is moving due south, which we can represent as a negative y-direction. - The passenger is walking toward the rear of the ship, which is also in the negative y-direction (since the ship is moving south). ### Step 3: Calculate the relative velocity of the passenger. The velocity of the passenger relative to the ground can be calculated by subtracting the passenger's speed from the liner's speed (both in the negative y-direction): \[ v_{\text{passenger, ground}} = v_{\text{liner}} - v_{\text{passenger}} = 5 \, \text{m/s} - 3.0 \, \text{m/s} = 2 \, \text{m/s} \] ### Conclusion The passenger's velocity relative to the ground is \( 2 \, \text{m/s} \) due south.