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 $20 m s^{- 1}$ . Find the average speed of the image after a time interval of 2 seconds?

10 m s^{- 1}

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20 m s^{- 1}

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2.5 m s^{- 1}

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5 m s^{- 1}

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Solution

The correct option is C $2.5 m s^{- 1}$
Given:
The distance of object $u$ from the pole is 60 m,
Radius of curvature $(r) = 40 m$
Let $f$ be the focal length and $v$ be the image distance
We know that,
$f = \frac{r}{2} = \frac{1}{2} \times 40 = 20 m$ .
Applying the sign conventions, $u = - 60 m$ and $f = + 20 m$
We know, $\frac{1}{v}$ + $\frac{1}{u}$ = $\frac{1}{f}$
$⟹ \frac{1}{v}$ - $\frac{1}{60}$ = $\frac{1}{20}$
$⟹ v = 15 m$
Positive sign shows that the image is virtual. Speed of the object $= 20 m s^{- 1}$ and time taken $= 2 s$
$∴$ Distance $=$ Speed $\times$ time
$⟹ 20 \times 2 = 40 m$
Therefore, now the position of object will be $u^{'} = 60 - 40 = 20 m$
Let $v^{'}$ be the new position of the image
Applying it in the mirror formula we get,
$\frac{1}{v^{'}}$ + $\frac{1}{u^{'}}$ = $\frac{1}{f}$
$⟹ \frac{1}{v^{'}}$ - $\frac{1}{20}$ = $\frac{1}{20}$
$⟹ \frac{1}{v^{'}}$ = $\frac{40}{20 \times 20}$
$⟹ v^{'} = 10 m$
So, the image moves from $15 m$ to $10 m$ in $2 s$ . Hence, its average speed = $\frac{15 - 10}{2} = 2.5 m s^{- 1}$

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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−1. Find the average speed of the image after a time interval of 2 seconds?

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 $20 m s^{- 1}$ . Find the average speed of the image after a time interval of 2 seconds?