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 refrence)

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 speeds
- The speed of the particle moving towards the mirror, \( v_p = 4 \, \text{cm/s} \).
- The speed of the observer moving in the same direction as the particle, \( v_o = 2 \, \text{cm/s} \).
- The speed of the mirror moving in the opposite direction, \( v_m = 10 \, \text{cm/s} \).

### Step 2: Determine the speed of the image with respect to the mirror
The speed of the image with respect to the mirror can be calculated using the formula:
\[ v_{i/m} = -v_p + v_m \]
Here, \( v_p \) is positive as it is moving towards the mirror, and \( v_m \) is negative because it is moving in the opposite direction.

Substituting the values:
\[ v_{i/m} = -4 \, \text{cm/s} + 10 \, \text{cm/s} = 6 \, \text{cm/s} \]
This means the image moves at \( 6 \, \text{cm/s} \) with respect to the mirror.

### Step 3: Find the speed of the image with respect to the observer
Now, we need to find the speed of the image with respect to the observer. The formula is:
\[ v_{i/o} = v_{i/m} - v_o \]
Substituting the values we have:
\[ v_{i/o} = 6 \, \text{cm/s} - 2 \, \text{cm/s} = 4 \, \text{cm/s} \]
However, since the image is moving in the opposite direction to the observer, we need to consider the direction. Therefore, we can express this as:
\[ v_{i/o} = -4 \, \text{cm/s} \]

### Step 4: Final speed of the image
The speed of the image observed by the observer is \( 4 \, \text{cm/s} \) in the leftward direction.

### Summary
The speed of the image observed by the observer is \( 4 \, \text{cm/s} \) towards the left.

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