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(A) For a convex mirror,the focal length $f = R/2 = 20/2 = 10 \ m$. The mirror formula is $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Case $1$: When ima

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(A) For a convex mirror,the focal length $f = R/2 = 20/2 = 10 \ m$. The mirror formula is $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Case $1$: When image position $v_1 = 25/3 \ m$ (virtual,so $v_1 = +25/3 \ m$):$\frac{1}{10} = \frac{3}{25} + \frac{1}{u_1} \implies \frac{1}{u_1} = \frac{1}{10} - \frac{3}{25} = \frac{5-6}{50} = -\frac{1}{50} \implies u_1 = -50 \ m$.Case $2$: When image position $v_2 = 50/7 \ m$ (virtual,so $v_2 = +50/7 \ m$):$\frac{1}{10} = \frac{7}{50} + \frac{1}{u_2} \implies \frac{1}{u_2} = \frac{1}{10} - \frac{7}{50} = \frac{5-7}{50} = -\frac{2}{50} = -\frac{1}{25} \implies u_2 = -25 \ m$.The displacement of the object is $\Delta u = u_2 - u_1 = -25 - (-50) = 25 \ m$.The time taken is $\Delta t = 30 \ s$.The speed of the object is $v_{obj} = \frac{|\Delta u|}{\Delta t} = \frac{25}{30} = \frac{5}{6} \ m/s$.Converting to $km/h$: $v_{obj} = \frac{5}{6} 	imes \frac{18}{5} = 3 \ km/h$.

An image of an object approaching a convex mirror of radius of curvature $20 \ m$ along its optical axis is observed to move from $\frac{25}{3} \ m$ to $\frac{50}{7} \ m$ in $30 \ s$. What is the speed of the object in $km/h$?

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