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(A) The mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. For a convex mirror,$R = +20 \,m$,so $f = +10 \,m$.Case $1$: Image position $v_1

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(A) The mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$. For a convex mirror,$R = +20 \,m$,so $f = +10 \,m$.Case $1$: Image position $v_1 = \frac{50}{7} \,m$. $\frac{1}{u_1} = \frac{1}{f} - \frac{1}{v_1} = \frac{1}{10} - \frac{7}{50} = \frac{5-7}{50} = -\frac{2}{50} = -\frac{1}{25} \implies u_1 = -25 \,m$.Case $2$: Image position $v_2 = \frac{25}{3} \,m$. $\frac{1}{u_2} = \frac{1}{f} - \frac{1}{v_2} = \frac{1}{10} - \frac{3}{25} = \frac{5-6}{50} = -\frac{1}{50} \implies u_2 = -50 \,m$.Distance traveled by object $\Delta u = |u_2 - u_1| = |-50 - (-25)| = 25 \,m$.Time taken $\Delta t = 30 \,s$.Speed of object $v = \frac{\Delta u}{\Delta t} = \frac{25}{30} \,m/s = \frac{5}{6} \,m/s$.Converting to $km/h$: $v = \frac{5}{6} 	imes \frac{18}{5} = 3 \,km/h$.

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

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