Find the velocity of the image of object $O$ with respect to ground in the given figure. The velocities of the object and the mirror w.r.t ground are given.

The correct option is C $\frac{7}{4}\text{cm/s}$ away from the mirror
For the given concave mirror,
$u=-30\text{cm},f=-10\text{cm}$
Relative velocity of object with respect to mirror,
$v_{om}=v_{og}-v_{mg}=2-(-1)=3\text{cm/s}$
Using mirror formula,
$\frac{1}{v}-\frac{1}{30}=\frac{-1}{10}$
$\Rightarrow \frac{1}{v}=\frac{-1}{15}$
$\Rightarrow v=-15\text{cm}$
So, $m=\frac{-v}{u}=\frac{-(-15)}{-30}=-\frac{1}{2}$
Also, velocity of the image when the object is moving along the principal axis,
$v_{im}=-m^2{v}_{om}$
$\Rightarrow v_{im}=-{(-\frac{1}{2})}^2\times 3=-\frac{3}{4}\text{cm/s}$
Now, the velocity of the image with respect to ground,
$v_{ig}=v_{im}+v_{mg}$
$\Rightarrow v_{ig}=-\frac{3}{4}-1=-\frac{7}{4}\text{cm/s}$
Since, the image is formed on the left side of the mirror $(v=-ve)$ and the image is moving in the left direction $(v_{im}=-ve)$, we can say that velocity of the image is away from the mirror.
Hence, the velocity of the image with respect to ground will be $\frac{7}{4}\text{cm/s}$ away from the mirror.
$\begin{array}{c}\text{Why this question ?}\\ \text{It provides unique blend of concept of relative}\\ \text{motion and mirror formula.}\\ \text{Tips: Apply the formula of image velocity w.r.t mirror}\\ \text{and be cautious in sign convention.}\\ \end{array}$