Consider the situation shown in the figure. The elevator is...

In the frame of the elevator, the effective acceleration due to gravity is:

$g_{\text{eff}} = g + a = 10 + 2.00 = 12.0 \text{ m s}^{-2}$

Let $a_{\text{rel}}$ be the acceleration of the blocks relative to the elevator. Writing the equations of motion for blocks A and B:

For block A: $T = m a_{\text{rel}}$

For block B: $m g_{\text{eff}} - T = m a_{\text{rel}}$

Adding the two equations, we get:

$m g_{\text{eff}} = 2m a_{\text{rel}} \Rightarrow a_{\text{rel}} = \dfrac{g_{\text{eff}}}{2} = \dfrac{12.0}{2} = 6.00 \text{ m s}^{-2}$

The distance fallen by block B relative to the elevator in time $t = 0.200 \text{ s}$ is:

$s = \dfrac{1}{2} a_{\text{rel}} t^2 = \dfrac{1}{2} (6.00) (0.200)^2 = 0.12 \text{ m} = 12.0 \text{ cm}$

The initial distance of block B from the mirror is $42.0 \text{ cm}$. At $t = 0.200 \text{ s}$, its distance from the mirror becomes:

$d = 42.0 - 12.0 = 30.0 \text{ cm}$

Applying the sign convention, the object distance is $u = -30.0 \text{ cm}$.

The mirror shown in the figure bulges towards the object, so it is a convex mirror. Its focal length is $f = +12.0 \text{ cm}$.

Using the mirror formula:

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

$\dfrac{1}{v} + \dfrac{1}{-30.0} = \dfrac{1}{12.0}$

$\dfrac{1}{v} = \dfrac{1}{12.0} + \dfrac{1}{30.0} = \dfrac{5 + 2}{60.0} = \dfrac{7}{60.0}$

$v = \dfrac{60.0}{7} \approx 8.57 \text{ cm}$

The distance between the image of block B and the mirror is $8.57 \text{ cm}$.