Car $$B$$ overtakes another car $$A$$ at a relative speed of 40 m s$$^{-1}$$. How fast will the image of car $$B$$ appear to move in the mirror of focal length 10 cm fitted in car $$A$$, when the car $$B$$ is 1.9 m away from the car $$A$$?

We start with the mirror formula for a spherical mirror, which in Cartesian sign convention is stated as

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

where
$$f$$ = focal length of the mirror,
$$u$$ = object distance from the mirror (measured along the incident-light direction and therefore negative for a real object in front of a convex mirror),
$$v$$ = image distance from the mirror (positive for a virtual image behind a convex mirror).

The side-view mirror on car A is a convex mirror whose focal length is given as 10 cm, so

$$f \;=\; +10\ \text{cm}.$$

Car B is 1.9 m behind car A, i.e.

$$u \;=\; -1.9\ \text{m} \;=\; -190\ \text{cm}.$$

Substituting these values into the mirror formula, we have

$$\frac{1}{10} \;=\;\frac{1}{v} + \frac{1}{-190}.$$

Rewriting,

$$\frac{1}{v} \;=\;\frac{1}{10} + \frac{1}{190} \;=\;\frac{19 + 1}{190} \;=\;\frac{20}{190} \;=\;\frac{2}{19}.$$

So

$$v \;=\;\frac{19}{2}\ \text{cm} \;=\; 9.5\ \text{cm}.$$

Next we require the speed with which the image distance $$v$$ is changing when car B is closing in on car A at 40 m s-1. First, convert this relative speed to centimetres per second:

$$40\ \text{m s}^{-1} \;=\; 40 \times 100\ \text{cm s}^{-1} \;=\; 4000\ \text{cm s}^{-1}.$$

Because car B is approaching the mirror, the object distance $$u$$ is decreasing, so

$$\frac{du}{dt} \;=\; -4000\ \text{cm s}^{-1}.$$

We differentiate the mirror formula with respect to time:

$$\frac{d}{dt}\!\left(\frac{1}{v} + \frac{1}{u}\right) = \frac{d}{dt}\!\left(\frac{1}{f}\right).$$

Since $$f$$ is constant, its derivative is zero, giving

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

Rearranging for $$\displaystyle\frac{dv}{dt}$$, we obtain

$$\frac{dv}{dt} \;=\;\left(\frac{v^{2}}{u^{2}}\right)\frac{du}{dt}.$$

Now substitute the numerical values:

$$\left(\frac{v^{2}}{u^{2}}\right) \;=\;\left(\frac{9.5^{2}}{190^{2}}\right) \;=\;\left(\frac{9.5}{190}\right)^{2} \;=\;(0.05)^{2} \;=\;0.0025.$$

Therefore

$$\frac{dv}{dt} \;=\;0.0025 \times (-4000) \;=\;-10\ \text{cm s}^{-1}.$$

The negative sign merely means the image is moving towards the mirror; its speed is the magnitude

$$|\tfrac{dv}{dt}| \;=\;10\ \text{cm s}^{-1} \;=\;0.1\ \text{m s}^{-1}.$$

Hence, the correct answer is Option A.