An object and a concave mirror of focal length f = 10cm both move...

Magnification expression for concave mirror is given by,

$m = \dfrac{f}{f - u}$

$= \dfrac{\left(- 10\right)}{\left(- 10\right) - \left(- 30\right)} = \dfrac{- 1}{2}$

For concave mirror, we can write,

${v}_{i, m} = - {m}^{2}{v}_{o . m}$

$⇒ \left({v}_{o} - {v}_{m}\right) \times {m}^{2} + \left({v}_{i} - {v}_{m}\right) = 0$

As given in the question, ${v}_{i} = 0$. Therefore,

$⇒ \left({v}_{o} - {v}_{m}\right) \times \dfrac{1}{4} = {v}_{m}$

$⇒ {v}_{o} = 5{v}_{m}$

$⇒ {v}_{m} = \dfrac{{v}_{o}}{5}$

$= \dfrac{15}{5}$

$= 3cm{s}^{- 1}$