Assume that you are sitting in a car. You see a person in the rear view mirror of radius of curvature 2m running towards you at t = 0. If person is running with velocity 5m/s, and it is at 9m distance from mirror at this instant. The average velocity of the image of man in first second is found to be 10x cm/s. Find the value of x.

From lens formula we know that

$v=\frac{fu}{u-f}$

Initial object distance $u=-9m$

Focal length of the mirror $f=\frac{R}{2}=\frac{2}{2}=1m$

Thus we get initial image distance $v_1=\frac{1\times (-9)}{-9-1}=\frac{9}{10}m$

After $1sec$, new object distance $u=-(9-5)=-4m$

Thus new image distance $v_2=\frac{1\times (-4)}{-4-1}=\frac{4}{5}m$

We get change in image distance $\Delta v=\frac{9}{10}-\frac{4}{5}=\frac{1}{10}m$

Average velocity $<v>=\frac{\Delta v}{\Delta t}=\frac{1}{10}m/s=10cm/s$

which implies value of $x$ is $1$.