A mirror of radius of curvature 20 cm and an object which is placed at a distance of 15 cm are both moving with velocities  $1m{s}^{-1}$  and  $10m{s}^{-1}$ as shown in  figure. The velocity of image at this situation is________   $(inm-s^{-1})$

$$\begin{gathered} U\sin g \frac{1}{v}+\frac{1}{u}=\frac{2}{R}, we get \\ \frac{1}{v}-\frac{1}{15}=-\frac{1}{10}\Rightarrow v=-30cm \\ Now, u\sin g v_{im}=-\frac{v^2}{u^2}{v}_{OM} \\ (v_i-v_m)=-\frac{v^2}{u^2}(v_0-v_m) \\ \Rightarrow v_i-\left(1\right)=-\frac{{(-30)}^2}{{(-15)}^2}[(-10)-\left(1\right)]\Rightarrow v_i=45cm{s}^{-1} \end{gathered}$$

So, the image will  move with velocity $45cm{s}^{-1}$