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A car is fitted with a convex side-view mirror of focal length $20 \mathrm{~cm}$. A second car $2.8 \mathrm{~m}$ behind the first car is overtaking the first car at relative speed of $15 \mathrm{~m} / \mathrm{s}$. The speed of the image of the second car as seen in the mirror of the first one is :
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The correct answer is:
$\frac{1}{15} \mathrm{~m} / \mathrm{s}$
$\frac{1}{15} \mathrm{~m} / \mathrm{s}$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$-\frac{1}{v^2} \frac{d v}{d t}-\frac{1}{u^2} \frac{d u}{d t}=0$
$\frac{d v}{d t}=-\frac{v^2}{u^2}\left(\frac{d u}{d t}\right)$
$f=20 \mathrm{~cm}$
$\frac{1}{u}+\frac{1}{-280}=\frac{1}{20}$
$\Rightarrow v=\frac{280}{15} \mathrm{~cm}$
$v_1=-\left(\frac{280}{15 \times 280}\right)^2 \times 15$
$=\frac{1}{15} \mathrm{~m} / \mathrm{s}$
$-\frac{1}{v^2} \frac{d v}{d t}-\frac{1}{u^2} \frac{d u}{d t}=0$
$\frac{d v}{d t}=-\frac{v^2}{u^2}\left(\frac{d u}{d t}\right)$
$f=20 \mathrm{~cm}$
$\frac{1}{u}+\frac{1}{-280}=\frac{1}{20}$
$\Rightarrow v=\frac{280}{15} \mathrm{~cm}$
$v_1=-\left(\frac{280}{15 \times 280}\right)^2 \times 15$
$=\frac{1}{15} \mathrm{~m} / \mathrm{s}$
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