The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature $R = 2\ \text{m}$. Another car approaches him from behind with a uniform speed of $90\ \text{km/h}$. When the car is at a distance of $24\ \text{m}$ from him, the magnitude of the acceleration of the image of the car in the side view mirror is $a$. The value of $100a$ is ______ $\text{m/s}^2$.
Numerical value type. Enter your answer.
Answer: 8
Focal length of the convex mirror $f = R/2 = 1\ \text{m}$. Let the object distance be $x$ (magnitude). The image distance behind the mirror is
$$v = \frac{fx}{x+f}$$
Differentiating with respect to time (with $\dot x$ constant):
$$\dot v = \frac{f^2}{(x+f)^2}\dot x, \qquad \ddot v = -\frac{2f^2}{(x+f)^3}\dot x^2$$
With $f = 1\ \text{m}$, $x = 24\ \text{m}$, $\dot x = 90\ \text{km/h} = 25\ \text{m/s}$:
$$a = \frac{2\times1\times625}{25^3} = \frac{1250}{15625} = 0.08\ \text{m/s}^2$$
$100a = 8$
Solution by Sreeraj P, M.Sc Physics