Question

Find the velocity of image w.r.t ground. The mirror is at rest and the object is moving perpendicular to the axis with $5\text{m/s}$ as in figure.The magnitude of velocity of image(in $\text{m/s}$) is

Answer 1

Magnification is $m=\frac{f}{f-u}=\frac{(-20)}{(-20)-(-30)}=-2$ and
$v=-mu=-(-2)(-30)=-60\text{cm}$
For the velocity component along the principal axis:
$(V_{IM}{)}_{\parallel }=-\frac{v^2}{u^2}(V_{OM}{)}_{\parallel }\Rightarrow (V_{IM}{)}_{\parallel }=0$
[Since $(V_{OM}{)}_{\parallel }=0]$
For the velocity component perpendiuclar to the principle axis:
$(V_{IM}{)}_{\perp }=m(V_{OM}{)}_{\perp }=m\frac{d{h}_o}{dt}\Rightarrow (V_{IM}{)}_{\perp }=(-2\left)\right(5)=-10\text{m/s}$
$\overrightarrow{V_{IG}}=\overrightarrow{V_{IM}}+\overrightarrow{V_{MG}}\Rightarrow V_{IG}=-10+0=-10\text{m/s}$ i.e. $10\text{m/s}$ (moving downwards)

Why this Question? Caution: All distances and velocities of object/image found using mirror formula are taken relative to pole of the mirror.