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 m. Another car approaches him from behind with a uniform speed of 90 km/hr. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the car in the side view mirror is ' a '. The value of 100 a is ________ $$m/s^{2}$$.

We are given a convex mirror with radius of curvature $$R = 2$$ m, so the focal length is $$f = +\frac{R}{2} = +1$$ m (positive for convex mirror by sign convention).

An object (car) approaches from behind at uniform speed $$v_0 = 90$$ km/hr $$= 25$$ m/s. When the car is at a distance of 24 m, the object distance is $$u = -24$$ m (negative by sign convention, since the object is in front of the mirror).

Using the mirror formula: $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{1} - \frac{1}{-24} = 1 + \frac{1}{24} = \frac{25}{24}$$

So $$v = \frac{24}{25}$$ m (positive, confirming virtual image behind the mirror).

To find the acceleration of the image, we use the relation between image position and object position. From the mirror formula:

$$v = \frac{uf}{u - f} = \frac{u \cdot 1}{u - 1} = \frac{u}{u - 1}$$

Differentiating $$v$$ with respect to $$u$$:

$$\frac{dv}{du} = \frac{(u - 1) - u}{(u - 1)^2} = \frac{-1}{(u - 1)^2}$$

Differentiating again:

$$\frac{d^2v}{du^2} = \frac{2}{(u - 1)^3}$$

The velocity of the image is:

$$\frac{dv}{dt} = \frac{dv}{du} \cdot \frac{du}{dt}$$

The acceleration of the image is:

$$\frac{d^2v}{dt^2} = \frac{d^2v}{du^2} \left(\frac{du}{dt}\right)^2 + \frac{dv}{du} \cdot \frac{d^2u}{dt^2}$$

Since the object moves with uniform speed, $$\frac{d^2u}{dt^2} = 0$$. The object approaches the mirror, so $$u$$ increases from $$-24$$ toward $$0$$. Therefore $$\frac{du}{dt} = +25$$ m/s.

At $$u = -24$$:

$$\frac{d^2v}{du^2} = \frac{2}{(-24 - 1)^3} = \frac{2}{(-25)^3} = \frac{2}{-15625} = -\frac{2}{15625}$$

$$\frac{d^2v}{dt^2} = -\frac{2}{15625} \times (25)^2 = -\frac{2 \times 625}{15625} = -\frac{1250}{15625} = -\frac{2}{25}$$

The magnitude of the acceleration of the image is:

$$|a| = \frac{2}{25}$$ m/s$$^2$$

Therefore:

$$100a = 100 \times \frac{2}{25} = 8$$

The answer is $$8$$.