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(D) For a convex mirror,the radius of curvature $R = 20 \ m$,so the focal length $f = +10 \ m$. The mirror formula is $1/v + 1/u = 1/f$. Given $v_1 =

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(D) For a convex mirror,the radius of curvature $R = 20 \ m$,so the focal length $f = +10 \ m$. The mirror formula is $1/v + 1/u = 1/f$. Given $v_1 = 25/3 \ m$ and $v_2 = 50/7 \ m$. For $v_1 = 25/3$: $1/u_1 = 1/10 - 3/25 = (5-6)/50 = -1/50$,so $u_1 = -50 \ m$. For $v_2 = 50/7$: $1/u_2 = 1/10 - 7/50 = (5-7)/50 = -2/50 = -1/25$,so $u_2 = -25 \ m$. The distance traveled by the object is $\Delta u = |u_2 - u_1| = |-25 - (-50)| = 25 \ m$. The time taken is $t = 30 \ s$. The speed of the object is $v_{obj} = \Delta u / t = 25 / 30 = 5/6 \ m/s$. To convert to $km/hr$,multiply by $18/5$: $v_{obj} = (5/6) 	imes (18/5) = 3 \ km/hr$.

The image of an object approaching a convex mirror of radius of curvature $20 \ m$ along its optical axis is observed to move from $25/3 \ m$ to $50/7 \ m$ in $30 \ s$. The speed of the object in $km/hr$ is

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