Question

An object is placed in front of a convex mirror at 60 m from the mirror. Its radius of curvature is 40 m. The object starts moving towards the mirror at 20 m/s. Then the average speed of the image after a time interval of 2 sec is ___ m/s.

Answer 1

The distance of object from the pole is 60 m and focal length is $\frac{1}{2}\times centreofcurvature=\frac{1}{2}\times 40=20m$

Applying sign conventions, u = -60, f = +20, v = ?

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}-\frac{1}{60}=\frac{1}{20}$

v = 15 cm. Positive sign shows that the image is virtual.

Now after 2 sec, the distance covered by object is $speed\times time=20\times 2=40m$

The distance of the object from the pole is 20 m.

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}-\frac{1}{20}=\frac{1}{20}$

$\frac{1}{v}=\frac{40}{20\times 20}$

v = 10 m

So, image moves from 15 m to 10 m in 2 sec.

Hence, $averagespeed=\frac{5}{2}=2.5m/s$