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 $$ 3.0 \mathrm{~m} / \mathrm{s} $$.

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Let the positive $$ y $$-direction represent south.

1. Convert the ship’s speed from kilometers per hour to meters per second:
   $$
   18\,\text{km/h} = 18 \times \frac{1000\,\text{m}}{3600\,\text{s}} = 5\,\text{m/s}
   $$
   Thus, the ship’s velocity is
   $$
   v_{\text{ship}} = 5\,\text{m/s} \quad (\text{south})
   $$

2. The passenger walks toward the rear of the ship at $$3.0\,\text{m/s}$$. Since the ship is moving south, the rear of the ship is pointing north. Therefore, the passenger’s velocity relative to the ship is
   $$
   v_{\text{passenger/ship}} = 3.0\,\text{m/s} \quad (\text{north})
   $$
   In our coordinate system where south is positive, north is negative:
   $$
   v_{\text{passenger/ship}} = -3.0\,\text{m/s}
   $$

3. The passenger’s velocity relative to the Earth (ground) is the sum of the ship’s velocity and the passenger’s velocity relative to the ship:
   $$
   v_{\text{passenger/Earth}} = v_{\text{ship}} + v_{\text{passenger/ship}} = 5\,\text{m/s} + (-3.0\,\text{m/s}) = 2\,\text{m/s}
   $$
   Since the result is positive, the direction is south.

Thus, the passenger’s velocity relative to the Earth is
$$
\boxed{2\,\text{m/s} \text{ (south)}}
$$
The passenger is moving at 2 meters per second south relative to the Earth.

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

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