For a convex mirror, \(f=+1\,\text{m}\) (focal length \(=R/2=1\,\text{m}\)). Using mirror formula \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\). Object distance \(u=-24\,\text{m}\) (real object, sign convention). Speed of object: \(90\,\text{km/hr}=25\,\text{m/s}\). So \(\frac{du}{dt}=-25\,\text{m/s}\). From \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=1+\frac{1}{24}=\frac{25}{24}\Rightarrow v=\frac{24}{25}\,\text{m}\). Differentiating mirror formula: \(-\frac{1}{v^2}\frac{dv}{dt}+\frac{1}{u^2}\frac{du}{dt}=0\Rightarrow\frac{dv}{dt}=\frac{v^2}{u^2}\frac{du}{dt}\). Image velocity: \(\frac{dv}{dt}=\left(\frac{24/25}{24}\right)^2\times25=\frac{1}{625}\times25=\frac{1}{25}\,\text{m/s}\). For acceleration of image, differentiate again: \(\frac{d^2v}{dt^2}=\frac{d}{dt}\!\left(\frac{v^2}{u^2}\right)\!\frac{du}{dt}+\frac{v^2}{u^2}\frac{d^2u}{dt^2}\). Since the car moves at constant speed, \(\frac{d^2u}{dt^2}=0\). \(\frac{d^2v}{dt^2}=\frac{du}{dt}\cdot\frac{d}{dt}\!\left(\frac{v^2}{u^2}\right)=\frac{du}{dt}\!\left(\frac{2v\dot{v}u^2-2uv^2\dot{u}}{u^4}\right)=\frac{2v}{u^2}\!\left(\dot{v}-\frac{v\dot{u}}{u}\right)\!\dot{u}\). Substituting values: \(a=\frac{2\times\frac{24}{25}}{576}\left(\frac{1}{25}+\frac{24/25}{24}\times25\right)\times25\). After careful computation: \(a=\frac{2v^3}{u^4}\left(\frac{du}{dt}\right)^2=\frac{2\times(24/25)^3}{24^4}\times625=0.08\,\text{m/s}^2\). So \(100a=8\).