Rear view mirrors

$\frac{2}{R}$ = $\frac{1}{u}$ + $\frac{1}{v}$ ;

differentiating wrt to time t;

0 = - $\frac{u'}{u^2}$ - $\frac{v'}{v^2}$

By sign convention u = -2 m; R = +2 m; frm ist eqn we get, v = +2/3 m and u' = -20kmph
and finaly frm eqn 2nd, we get. v' = $\frac{20}{9}$ kmph

When car B is at a distance 2m from car A, we can find the position of image of B in the concave mirror.

$\frac{1}{v} + \frac{1}{-2} = \frac{2}{R}$

As R=2m,

$\frac{1}{v} + \frac{1}{-2} = \frac{2}{2}$

$\frac{1}{v} + \frac{1}{-2} = 1$

$v=\frac{2}{3}$

Now, relative velocity of B w.r.t A is 60 - 40 = 20kmph

Now, velocity of image in curved mirror = $-m_{transverse}^{2}\times(velocity of object)$

$m_{t} = \frac{-v}{u} = \frac{-2/3}{2} = \frac{-1}{3}$

$m_{t}^{2} = \frac{1}{9}$

Therefore, velocity of image is = -\frac{1}{9}\times20 kmph = 2.22222kmph