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 \(\mathrm{R}=2 \mathrm{~m}\). Another car approaches him from behind with a uniform speed of \(90 \mathrm{~km} / \mathrm{hr}\). 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 _______ \(\mathrm{m} / \mathrm{s}^2\).

\(\begin{aligned}
& \mathrm{v}=\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}}=\frac{-24 \cdot 1}{-24-1}=\frac{24}{25} \\
& \mathrm{~m}=\frac{-\mathrm{v}}{\mathrm{u}}=-\frac{24}{25(-24)}=\frac{1}{25} \\
& \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
& \mathrm{v}_{\mathrm{I}}=-\mathrm{m}^2 \cdot \mathrm{v}_0=\frac{-1}{(25)^2} \cdot 25=\frac{-1}{25}
\end{aligned}\)
Diff.
\(\begin{aligned}
& \frac{-1}{\mathrm{v}^2}\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)+\frac{1}{\mathrm{u}^2}\left(\frac{\mathrm{du}}{\mathrm{dt}}\right)=0\left[\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{i}} ; \frac{\mathrm{du}}{\mathrm{dt}}=\mathrm{v}_0\right] \\
& \frac{+2}{\mathrm{v}^3} \cdot\left(\mathrm{v}_{\mathrm{t}}\right)^2-\frac{1}{\mathrm{v}^2} \cdot \mathrm{a}_{\mathrm{L}}-\frac{2}{\mathrm{u}^3} \cdot\left(\mathrm{v}_0\right)^2+\frac{1}{\mathrm{u}^2} \cdot \mathrm{a}_0=0 \\
& \frac{\mathrm{a}_{\mathrm{I}}}{\mathrm{v}^2}=\frac{2}{\mathrm{v}^3} \cdot \mathrm{v}_{\mathrm{I}}^2-\frac{2}{\mathrm{u}^3} \cdot \mathrm{v}_0^2 \\
& \mathrm{a}_{\mathrm{I}}=\frac{2}{\mathrm{v}} \cdot \mathrm{v}_{\mathrm{I}}^2-\frac{2 \mathrm{v}^2}{\mathrm{u}^3} \cdot \mathrm{v}_0^2 \\
& =\frac{2 \cdot 25}{24} \cdot \frac{1}{25} \cdot \frac{1}{25}-\frac{2}{(24)^3} \cdot \frac{24}{25} \cdot \frac{24}{25} \cdot 25 \cdot 25 \\
& \mathrm{a}_{\mathrm{I}}=\frac{2}{24.25}-\frac{2}{24} \\
& \mathrm{a}_1=\frac{2}{24} \cdot \frac{-24}{25}=\frac{-2}{25} \\
& 100 a_1=\frac{2}{25} \times 100=8
\end{aligned}\)