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.

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Solution

The distance of object from the pole is 60 m and focal length is $\frac{1}{2} \times c e n t r e o f c u r v a t u r e = \frac{1}{2} \times 40 = 20 m$

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 $s p e e d \times t i m e = 20 \times 2 = 40 m$

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, $a v e r a g e s p e e d = \frac{5}{2} = 2.5 m / s$

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