A particle moves perpendicular towards a plane mirror with a constant speed of `4 cm//s`. What is the speed of the image observed by an observed moving with `2 cm//s` along the same direction? Mirror is also moving with a speed of `10cm//s` in the opposite direction.
(All speeds are with respect to ground frame of reference)

Options

A. `4 cm//s`
B. `12 cm//s`
C. `14 cm//s`
D. `26 cm//s`

Explanation

To solve the problem step by step, we need to analyze the motion of the particle, the mirror, and the observer.

### Step 1: Identify the velocities
- The particle moves towards the mirror at a speed of \(4 \, \text{cm/s}\).
- The observer moves in the same direction as the particle at a speed of \(2 \, \text{cm/s}\).
- The mirror moves in the opposite direction at a speed of \(10 \, \text{cm/s}\).

### Step 2: Calculate the speed of the image with respect to the ground
The speed of the image formed by the mirror can be calculated using the following relationship:
\[ \text{Velocity of image with respect to ground} = \text{Velocity of object with respect to ground} + \text{Velocity of mirror with respect to ground} \]
Since the mirror is moving in the opposite direction, we need to subtract its speed:
\[ \text{Velocity of image} = 4 \, \text{cm/s} + 10 \, \text{cm/s} = 14 \, \text{cm/s} \]

### Step 3: Calculate the speed of the image with respect to the observer
Now, we need to find the speed of the image relative to the observer. The observer is moving at \(2 \, \text{cm/s}\) in the same direction as the particle and the image. Therefore, we subtract the observer's speed from the image's speed:
\[ \text{Velocity of image with respect to observer} = \text{Velocity of image with respect to ground} - \text{Velocity of observer with respect to ground} \]
Substituting the values:
\[ \text{Velocity of image with respect to observer} = 14 \, \text{cm/s} - 2 \, \text{cm/s} = 12 \, \text{cm/s} \]

### Final Answer
The speed of the image observed by the observer is \(12 \, \text{cm/s}\).

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