A plane mirror is moving with a uniform speed $5{ \mathrm{ms}}^{-1}$ along negative x-direction and an observer P is moving with a velocity of $10{ \mathrm{ms}}^{-1}$along +x direction. Calculate the velocity of image of an object moving with a velocity of $10\sqrt{2}{ \mathrm{ms}}^{-1}$ as shown in the figure, as observed by the observer. Also find its magnitude and direction 

Let${\overrightarrow{v}}_O$ be the velocity of the object O, ${\overrightarrow{v}}_p$ be the velocity of the observer P, ${\overrightarrow{v}}_M$ be the velocity of the mirror and ${\overrightarrow{v}}_M$be the velocity of image (Assume all these velocities w.r.t. ground), then

$$\begin{gathered} {\overrightarrow{v}}_O=\frac{10\sqrt{2}}{\sqrt{2}}(\hat{i}+\hat{j}) \\ {\overrightarrow{v}}_O=10(\hat{i}+\hat{j}) \\ {\overrightarrow{v}}_P=10\hat{i} \\ {\overrightarrow{v}}_M=-5\hat{i} \end{gathered}$$

${\left({\overrightarrow{v}}_{IM}\right)}_{\perp }=-{\left({\overrightarrow{v}}_{OM}\right)}_{\perp }$, where the axis perpendicular to the mirror is the x-axis.

$$\begin{gathered} \Rightarrow {\left({\overrightarrow{v}}_I\right)}_x-{\left({\overrightarrow{v}}_M\right)}_x=-{\left({\overrightarrow{v}}_O\right)}_x+{\left({\overrightarrow{v}}_M\right)}_x \\ \Rightarrow {\left({\overrightarrow{v}}_I\right)}_x=2{\left({\overrightarrow{v}}_M\right)}_x-{\left({\overrightarrow{v}}_O\right)}_x \\ \Rightarrow {\left({\overrightarrow{v}}_I\right)}_x=2(-5\hat{i})-10\hat{i} \\ \Rightarrow {\left({\overrightarrow{v}}_I\right)}_x=-20\hat{i} \end{gathered}$$

Further, parallel to the mirror, i.e., along y-axis, have

${\left({\overrightarrow{v}}_I\right)}_y={\left({\overrightarrow{v}}_O\right)}_y=10\hat{j}$

Since ${\overrightarrow{v}}_I={\left({\overrightarrow{v}}_I\right)}_x+{\left({\overrightarrow{v}}_I\right)}_y$

So, absolute velocity of the image is

Now ${\overrightarrow{v}}_{iP}={\hat{v}}_i-{\overline{v}}_P$

$$\begin{gathered} \Rightarrow {\overrightarrow{v}}_{IP}=-20\hat{i}+10\hat{j}-10\hat{i} \\ \Rightarrow {\overrightarrow{v}}_{IP}=-30\hat{i}+10\hat{j} \\ \Rightarrow \left|{\overrightarrow{v}}_{IP}\right|=\sqrt{900+100}=10\sqrt{10}{ \mathrm{ms}}^{-1} \end{gathered}$$

If 

$$\begin{gathered} \beta is the anglemadeby{\overrightarrow{v}}_{IP}with-xaxis,then \tan \beta =\frac{10}{30} \\ \Rightarrow \beta ={\tan }^{-1}\left(\frac{1}{3}\right)\text{, with }-x\text{ axis } \end{gathered}$$