A car is moving towards large convex mirror of radius of curvature 20 m at speed of 4 m/s. When it is at 40 m from the mirror, speed of its image will be

2 m/s

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1.5 m/s

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0.16 m/s

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0.08 m/s

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Solution

The correct option is C 0.16 m/s
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
By differentiating on both sides,
$\frac{d v}{d t} = - (\frac{v^{2}}{u^{2}}) \frac{d u}{d t}$
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{10} = \frac{1}{v} - \frac{1}{40} \Rightarrow \frac{1}{10} + \frac{1}{40} = \frac{1}{v}$
$\frac{1}{v} = \frac{4 + 1}{40} \Rightarrow v = 8 m$
$\frac{d v}{d t} = - (\frac{8}{40})^{2} \times 4 = - \frac{4}{25} = - 0.16 m / s$
So the speed of the image is $0.16 m / s$ towards the mirror.

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A car is moving towards large convex mirror of radius of curvature 20 m at speed of 4 m/s. When it is at 40 m from the mirror, speed of its image will be

The correct option is C 0.16 m/s1f=1v+1u By differentiating on both sides, dvdt=−(v2u2)dudt 1f=1v+1u⇒110=1v−140⇒110+140=1v 1v=4+140⇒v=8 m dvdt=−(840)2×4=−425=−0.16 m/s So the speed of the image is 0.16 m/s towards the mirror.