Step-by-step Solution:

1. Establish Coordinate System and Sign Convention

We will use the standard Cartesian sign convention for spherical mirrors.

• The pole of the mirror is the origin (0,0).
• The principal axis is the x-axis.
• The direction of incident light is taken as the positive x-direction (usually from left to right).
• Distances measured in the direction of incident light are positive, and those measured opposite to it are negative.
• A concave mirror has a negative focal length.

2. List Given Parameters with Signs

• Focal length of the concave mirror, $f = -10 \mathrm{~cm}$.
• The distance of the object from the mirror is given as $u=30 \mathrm{~cm}$. According to our sign convention, since the object is placed in front of the mirror, the object distance is $u = -30 \mathrm{~cm}$.
• The object moves towards the mirror with a speed of $V_{0}=15 \mathrm{~cm} \mathrm{~s}^{-1}$ with respect to the lab frame. Since it moves towards the mirror, it moves in the positive x-direction. So, its velocity is $V_{o} = +15 \mathrm{~cm} \mathrm{~s}^{-1}$.
• The image is instantaneously at rest with respect to the laboratory frame. So, the velocity of the image is $V_{i} = 0 \mathrm{~cm} \mathrm{~s}^{-1}$.
• We need to find the velocity of the mirror, $V_{m}$.

3. Calculate the Image Position (v)

We use the mirror formula to find the position of the image: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ Substituting the given values: $\frac{1}{v} + \frac{1}{-30} = \frac{1}{-10}$ $\frac{1}{v} = -\frac{1}{10} + \frac{1}{30}$ $\frac{1}{v} = \frac{-3 + 1}{30} = -\frac{2}{30} = -\frac{1}{15}$ $v = -15 \mathrm{~cm}$ The negative sign indicates that the image is formed in front of the mirror, which means it is a real image, consistent with the problem statement.

4. Determine the Velocity Relationship

The relationship between the velocities of the object, image, and mirror along the principal axis can be found by differentiating the mirror formula with respect to time $t$: $\frac{d}{dt} \left( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \right)$ Since $f$ is constant for a given mirror: $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$ Here, $u$ and $v$ are positions relative to the moving mirror. Let the positions of the object, image, and mirror in the lab frame be $x_o, x_i, x_m$ respectively. Then $u = x_o - x_m$ and $v = x_i - x_m$. Their time derivatives are:

• $\frac{du}{dt} = \frac{d(x_o - x_m)}{dt} = V_o - V_m$ (velocity of object w.r.t. mirror)
• $\frac{dv}{dt} = \frac{d(x_i - x_m)}{dt} = V_i - V_m$ (velocity of image w.r.t. mirror)

Substituting these into the differentiated equation: $-\frac{1}{v^2}(V_i - V_m) - \frac{1}{u^2}(V_o - V_m) = 0$ $\frac{V_i - V_m}{v^2} = -\frac{V_o - V_m}{u^2}$ $V_i - V_m = -\left(\frac{v}{u}\right)^2 (V_o - V_m)$ The lateral magnification is $m = -\frac{v}{u}$. So, $\left(\frac{v}{u}\right)^2 = (-m)^2 = m^2$. The velocity relationship is: $V_i - V_m = -m^2 (V_o - V_m)$

5. Calculate Magnification (m)

$m = -\frac{v}{u} = -\frac{-15}{-30} = -\frac{1}{2}$ $m^2 = \left(-\frac{1}{2}\right)^2 = \frac{1}{4}$

6. Solve for the Mirror's Velocity (Vm)

Now, substitute all the known values into the velocity equation: $0 - V_m = -\frac{1}{4} (15 - V_m)$ $-V_m = -\frac{15}{4} + \frac{1}{4}V_m$ $-V_m - \frac{1}{4}V_m = -\frac{15}{4}$ $-\frac{5}{4}V_m = -\frac{15}{4}$ $5V_m = 15$ $V_m = 3 \mathrm{~cm} \mathrm{~s}^{-1}$

The magnitude of the velocity of the mirror is $3 \mathrm{~cm} \mathrm{~s}^{-1}$. The positive sign indicates that the mirror is moving in the +x direction (away from the object).

Final Answer: The magnitude of $V_m$ is 3.