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

Convex Mirror Problem Solution

Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ where focal length $f = R/2 = 10$ m (positive by convention).

For initial image position $v_1 = 25/3$ m:
$\frac{1}{u_1} = \frac{1}{10} - \frac{3}{25} = -\frac{1}{50}$
$u_1 = -50$ m

For final image position $v_2 = 50/7$ m:
$\frac{1}{u_2} = \frac{1}{10} - \frac{7}{50} = -\frac{1}{25}$
$u_2 = -25$ m

The object moves from $u_1 = -50$ m to $u_2 = -25$ m, covering distance $\Delta u = 25$ m in 30 seconds.

Speed is $\frac{25}{30} = \frac{5}{6}$ m/s. Converting to km/h:
$\frac{5}{6} \times \frac{18}{5} = 3$ km/h

The object's speed is $\boxed{3}$ km/h.