1. Interpret the situation

Car B is behind car A and is seen in the mirror fitted in car A. Since vehicles use a convex mirror as rear-view mirror, we take:

$f = +10\text{ cm} = 0.10\text{ m}$

The distance of car B from the mirror of car A is:

$u = -1.9\text{ m}$

(using Cartesian sign convention for mirrors, object in front of mirror has negative object distance).

The relative speed with which B approaches A is:

$\frac{du}{dt} = +40\text{ m s}^{-1}$

because $u$ is negative and its magnitude is decreasing, so $u$ increases toward zero.

2. Use the mirror formula

For a spherical mirror,

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

Differentiate w.r.t. time:

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

So,

$\frac{dv}{dt} = -\frac{v^2}{u^2}\frac{du}{dt}$

We need $v$ first.

3. Find the image distance

$\frac{1}{0.10} = \frac{1}{v} + \frac{1}{-1.9}$

$10 = \frac{1}{v} - \frac{1}{1.9}$

$\frac{1}{v} = 10 + \frac{1}{1.9} = 10 + 0.5263 = 10.5263$

$v \approx 0.095\text{ m}$

So,

$v = 9.5\text{ cm}$

4. Compute image speed

$\frac{dv}{dt} = -\frac{(0.095)^2}{(1.9)^2}(40)$

Now,

$\frac{0.095}{1.9} = 0.05$

Hence,

$\frac{v^2}{u^2} = (0.05)^2 = 0.0025$

Therefore,

$\frac{dv}{dt} = -(0.0025)(40) = -0.1\text{ m s}^{-1}$

Negative sign means the image moves toward the mirror. The speed is

$0.1\text{ m s}^{-1}$

5. Match with options

The correct option is:

D: $0.1\text{ m s}^{-1}$

6. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.