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(B) Given: $R = 2 \ m$,so focal length $f = R/2 = 1 \ m$. The speed of the object $v_0 = 90 \ km/h = 90 	imes (5/18) = 25 \ m/s$. Since the object is

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$8$

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(B) Given: $R = 2 \ m$,so focal length $f = R/2 = 1 \ m$. The speed of the object $v_0 = 90 \ km/h = 90 	imes (5/18) = 25 \ m/s$. Since the object is approaching,$u = -24 \ m$ and $du/dt = v_0 = 25 \ m/s$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we differentiate with respect to time: $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$,which gives $v_I = \frac{dv}{dt} = -(\frac{v}{u})^2 v_0 = -m^2 v_0$.For $u = -24 \ m$ and $f = 1 \ m$,$v = \frac{uf}{u-f} = \frac{(-24)(1)}{-24-1} = \frac{24}{25} \ m$.Magnification $m = -v/u = -(\frac{24/25}{-24}) = \frac{1}{25}$.Velocity of image $v_I = -m^2 v_0 = -(\frac{1}{25})^2 (25) = -\frac{1}{25} \ m/s$.Differentiating the velocity equation $\frac{dv}{dt} = -(\frac{v}{u})^2 \frac{du}{dt}$ again with respect to time:$a_I = \frac{d^2v}{dt^2} = -[2(\frac{v}{u})(\frac{u \frac{dv}{dt} - v \frac{du}{dt}}{u^2})] v_0 - (\frac{v}{u})^2 a_0$. Since $a_0 = 0$,$a_I = -2(\frac{v}{u})(\frac{u v_I - v v_0}{u^2}) v_0$.Substituting values: $a_I = -2(\frac{24/25}{-24})(\frac{(-24)(-1/25) - (24/25)(25)}{(-24)^2}) (25) = -2(-\frac{1}{25})(\frac{24/25 - 24}{576}) (25) = 2(\frac{1}{25})(\frac{24 - 600}{25 	imes 576}) (25) = 2(\frac{-576}{25 	imes 576}) = -\frac{2}{25} \ m/s^2$.The magnitude $a = |-\frac{2}{25}| = 0.08 \ m/s^2$. Thus,$100a = 100 	imes 0.08 = 8$.

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