A real object is placed in front of a convex mirror (fixed).The object is moving toward the mirror. If `v_(0)` is the speed of object and `v_(i)` is the speed of image, then

Options

A. `v_(i) lt v_(0)` always
B. `v_(i) gt v_(0)` always
C. `v_(i) gt v_(0)` initially and then `v_(0) gt v_(i)`
D. `v_(i) lt v_(0)` initially and then `v_(i) gt v_(0)`

Explanation

To solve the problem, we need to analyze the behavior of a real object moving towards a convex mirror and how it affects the speed of the image formed by the mirror.

### Step-by-Step Solution:

1. **Understanding the System**:
- A real object is placed in front of a convex mirror.
- The object is moving towards the mirror with a speed denoted as \( v_0 \).

2. **Image Formation in a Convex Mirror**:
- A convex mirror always forms a virtual image behind the mirror.
- As the object approaches the mirror, the image moves from the focus of the mirror towards the pole of the mirror.

3. **Speed of the Object and Image**:
- The speed of the object is \( v_0 \).
- We need to find the relationship between the speed of the image \( v_i \) and the speed of the object \( v_0 \).

4. **Analyzing the Motion**:
- As the object moves closer to the mirror, the distance between the object and the mirror decreases.
- The image formed by the convex mirror moves towards the pole but at a slower rate compared to the object.

5. **Conclusion on Speeds**:
- Since the image moves slower than the object as the object approaches the mirror, we can conclude that:
\[ v_i < v_0 \]
- Therefore, the speed of the image \( v_i \) is always less than the speed of the object \( v_0 \).

6. **Final Answer**:
- The correct option is that \( v_0 \) is greater than \( v_i \) always.