A train is moving slowly on a straightly track with a constant speed of `2 ms^(-1)` . A passenger in the train starts walking at a steady speed of `2 ms^(-1)` to the back of the train in the opposite direction of the motion of the train . So, to an observer standing on the platform directly in the front of that passenger appears to be

Options

A. `4ms ^(-1)`
B. `2ms^(-1)`
C. `2ms^(-1)` in the opposite direction of the train
D. zero

Answer

To solve the problem, we need to analyze the motion of the train and the passenger in relation to an observer on the platform.

### Step-by-Step Solution:

1. **Identify the velocities**:
- The train is moving with a constant speed of \(2 \, \text{m/s}\) in a positive direction (let's assume the positive direction is towards the right).
- The passenger is walking towards the back of the train (which is in the negative direction) at a speed of \(2 \, \text{m/s}\).

2. **Assign the velocities**:
- Let the velocity of the train \(v_t = +2 \, \text{m/s}\).
- Let the velocity of the passenger with respect to the train \(v_p = -2 \, \text{m/s}\).

3. **Calculate the velocity of the passenger with respect to the ground**:
- To find the velocity of the passenger with respect to the ground, we can use the formula:
\[ v_{pg} = v_t + v_p \]
- Substituting the values:
\[ v_{pg} = 2 \, \text{m/s} + (-2 \, \text{m/s}) = 2 \, \text{m/s} - 2 \, \text{m/s} = 0 \, \text{m/s} \]

4. **Interpret the result**:
- The velocity of the passenger with respect to the ground is \(0 \, \text{m/s}\). This means that to an observer standing on the platform, the passenger appears to be stationary.

5. **Conclusion**:
- Therefore, to an observer standing on the platform directly in front of the passenger, the passenger appears to be at rest (stationary).

### Final Answer:
The passenger appears to be stationary to an observer on the platform.

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