Image of an object approaching a convex mirror of radius of curvature 20m along its optical axis is observed to move from 25/3m to 50/7m in 30 seconds. What is the speed of the object in km per hour?

Given, R=20m, f=10m
From the mirror equation; 1/v + 1/u = 1/f
Initially, the image is at 25/3 m. So, 1/(25/3) + 1/u1 = 1/10
1/u1 = 1/10 - 3/25 = (5 - 6)/50 = -1/50
u1 = -50m
Furthermore, when the image of the object is at 50/7 m.
1/v2 + 1/u2 = 1/f
1/(50/7) + 1/u2 = 1/10
1/u2 = 1/10 - 7/50 = (5 - 7)/50 = -2/50 = -1/25
u2 = -25m
Change in the object's position = 50 - 25 = 25m
Speed = displacement/time = 25/30 = 5/6 m/s
Speed in km/hr = (5/6) * (18/5) = 3 km/hr