Given, R = 2m, So, f = 1m (as $f=\frac{R}{2}$)

$${V_0} = 90\,km/hr = 90 \times {{1000} \over {3600}} = 25\,m/s$$

$v_0$ is uniform, so $${a_0} = 0$$

$$u = - 24\,m$$

Using mirror formula,

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

$$ \Rightarrow {1 \over v} + {1 \over { - 24}} = {1 \over 1} \Rightarrow {1 \over v} = 1 + {1 \over {24}} = {{25} \over {24}}$$

$$ \Rightarrow v = {{24} \over {25}}m$$

$$m = {{ - v} \over u}$$

by differentiating mirror formula,

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

$$ \Rightarrow - {1 \over {{v^2}}}{v_I} - {1 \over {{u^2}}}{v_0} = 0$$ ...... (1)

$$ \Rightarrow {v_I} = {{ - {v^2}} \over {{u^2}}}{v_0} = - {{{{24}^2}} \over {{{25}^2}}}$$

$$ \Rightarrow {v_I} = {{ - 1} \over {25}}$$ m/s

by differentiating (1) w.r.t. time,

$${{ - 1} \over {{v^2}}}{a_i} + {2 \over {{v^3}}}v_I^2 - {1 \over {{u^2}}}{a_0} + {2 \over {{u^3}}}v_0^2 = 0$$

$${a_I} = {2 \over v}v_I^2 + {{2{v^2}} \over {{u^3}}}v_0^2$$

$$ = {2 \over {24}}{1 \over {{{25}^2}}} + {2 \over { - {{24}^3}}}{{{{24}^2}} \over {{{25}^2}}}{25^2}$$

$$ = {1 \over {12}}\left[ {{1 \over {25}} - 1} \right] = {1 \over {12}} \times {{ - 24} \over {25}} = - {2 \over {25}}$$

$$ \Rightarrow 100\,{a_I} = {{ - 2} \over {25}} \times 100 = 8\,m/{s^2}$$