In the figure shown consider the first reflection at the plane mirror and second at the convex mirror. AB is object. 

I^st image (A₁ B₁) is formed due to plane mirror to the left of it with magnification 1. It is virtual. This A₁B₁ acts as real object to the
concave mirror of focal length 
$\mathrm{f}=\frac{\mathrm{R}}{2}=\frac{120}{2}=60\mathrm{cm}$
 

A₂B₂ is the image of A₁B₁ formed by convex mirror. 
$$\begin{gathered} \mathrm{AB}={\mathrm{A}}_1{\mathrm{B}}_1=30\mathrm{cm} \\ \text{ For }{\mathrm{A}}_1:\mathrm{u}=-60\mathrm{cm},\mathrm{f}=+60\mathrm{cm},\mathrm{V}? \\ \frac{1}{\mathrm{V}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \frac{1}{\mathrm{V}}-\frac{1}{60} \\ =\frac{1}{60}\Rightarrow \mathrm{V}=+30\mathrm{cm}\left(\mathrm{P}\to {\mathrm{A}}_2\right) \end{gathered}$$
$$\begin{gathered} \text{ For }{\mathrm{B}}_1:\frac{1}{\mathrm{V}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \frac{1}{\mathrm{V}}-\frac{1}{90}=\frac{1}{60} \\ \Rightarrow \mathrm{V}=+36\mathrm{cm}\left(\mathrm{P}\to {\mathrm{B}}_2\right) \\ \text{ length of }{\mathrm{A}}_2{\mathrm{B}}_2=36-30=6\mathrm{cm} \\ \frac{{\mathrm{A}}_2{\mathrm{B}}_2}{\mathrm{AB}}=\frac{6}{30}\Rightarrow {\mathrm{A}}_2{\mathrm{B}}_2=\frac{\mathrm{AB}}{5} \end{gathered}$$
In the figure, it can be observed that both A-B 
and A₂–B₂ are left to right i.e., A₂B₂ is erected w.r.t AB. A₂B₂ is virtual 
Velocity of image formed by a mirror is always opposite to the object. 

If AB moves left$\Rightarrow {\mathrm{A}}_1{\mathrm{B}}_1\text{ moves right wards }\Rightarrow {\mathrm{A}}_2{\mathrm{B}}_2$ moves leftwards i.e., towards the mirror.