An 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 in 30 seconds. What is the speed of the object in km per hour?

Given:
• Radius of curvature of the convex mirror, R=20 mR = 20 \, \text{m}
• Initial image distance, v1=253 mv_1 = \frac{25}{3} \, \text{m}
• Final image distance, v2=507 mv_2 = \frac{50}{7} \, \text{m}
• Time duration, t=30 secondst = 30 \, \text{seconds}
We are asked to find the speed of the object in km per hour.
Step 1: Mirror Equation
For a convex mirror, the relationship between the object distance uu, the image distance vv, and the focal length ff is given by the mirror equation:
1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}
Also, the focal length ff of the mirror is related to the radius of curvature RR by:
f=R2f = \frac{R}{2}
Substituting R=20 mR = 20 \, \text{m}, we get:
f=202=10 mf = \frac{20}{2} = 10 \, \text{m}
Step 2: Initial and Final Object Distances
Since the image distance for a convex mirror is always positive and virtual, we can use the mirror equation to find the initial and final object distances.
For the initial image position v1=253 mv_1 = \frac{25}{3} \, \text{m}, using the mirror equation:
1f=1v1+1u1\frac{1}{f} = \frac{1}{v_1} + \frac{1}{u_1}
Substitute the known values:
110=1253+1u1\frac{1}{10} = \frac{1}{\frac{25}{3}} + \frac{1}{u_1}
Simplify:
110=325+1u1\frac{1}{10} = \frac{3}{25} + \frac{1}{u_1}
Rearranging to solve for u1u_1:
1u1=110−325\frac{1}{u_1} = \frac{1}{10} - \frac{3}{25}
Finding a common denominator:
1u1=550−650=−150\frac{1}{u_1} = \frac{5}{50} - \frac{6}{50} = \frac{-1}{50}
Thus,
u1=−50 mu_1 = -50 \, \text{m}
Now, for the final image position v2=507 mv_2 = \frac{50}{7} \, \text{m}, using the same mirror equation:
110=1507+1u2\frac{1}{10} = \frac{1}{\frac{50}{7}} + \frac{1}{u_2}
Simplify:
110=750+1u2\frac{1}{10} = \frac{7}{50} + \frac{1}{u_2}
Rearranging to solve for u2u_2:
1u2=110−750\frac{1}{u_2} = \frac{1}{10} - \frac{7}{50}
Finding a common denominator:
1u2=550−750=−250\frac{1}{u_2} = \frac{5}{50} - \frac{7}{50} = \frac{-2}{50}
Thus,
u2=−25 mu_2 = -25 \, \text{m}
Step 3: Calculate the Speed of the Object
The object moves from u1=−50 mu_1 = -50 \, \text{m} to u2=−25 mu_2 = -25 \, \text{m} in 30 seconds. The displacement of the object is:
Δu=u2−u1=−25−(−50)=25 m\Delta u = u_2 - u_1 = -25 - (-50) = 25 \, \text{m}
The speed of the object is the rate of change of distance, so:
Speed=ΔuΔt=2530=56 m/s\text{Speed} = \frac{\Delta u}{\Delta t} = \frac{25}{30} = \frac{5}{6} \, \text{m/s}
Step 4: Convert Speed to km/h
To convert the speed from m/s to km/h, multiply by 3.6:
Speed in km/h=56×3.6=3 km/h\text{Speed in km/h} = \frac{5}{6} \times 3.6 = 3 \, \text{km/h}
Final Answer:
The speed of the object is 3 km/h.