A particle travels at a constant speed V from a large distance...

The mirror formula is given by:

$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

Differentiating with respect to time $t$:

$-\dfrac{1}{v^2} \dfrac{dv}{dt} - \dfrac{1}{u^2} \dfrac{du}{dt} = 0$

$\dfrac{dv}{dt} = -\left(\dfrac{v}{u}\right)^2 \dfrac{du}{dt}$

The speed of the object is $V$, so \left|\dfrac{du}{dt}\right| = V$. The speed of the image is given by$v_i = \left|\dfrac{dv}{dt}\right| = m^2 V$, where$m = -\dfrac{v}{u}$is the transverse magnification.<br><br>From the mirror formula, the magnification can be expressed as:<br><br>$m = \dfrac{f}{f - u}$<br><br>For a concave mirror, the focal length is$f = -\dfrac{R}{2}$and the object distance is$u = -x$. Substituting these values:<br><br>$m = \dfrac{-\dfrac{R}{2}}{-\dfrac{R}{2} - (-x)} = \dfrac{-\dfrac{R}{2}}{x - \dfrac{R}{2}} = \dfrac{-R}{2x - R}$<br><br>Therefore, the speed of the image is:<br><br>$v_i = m^2 V = \left(\dfrac{-R}{2x - R}\right)^2 V = \dfrac{R^2 V}{(2x - R)^2}