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A car is fitted with a convex side -view mirror of focal length 20cm. A second car 2.8m behind the first car is overtaking the first car at relative speed of $15 m/s$ . The speed of the image of the second car as seen in the mirror of the first one is :

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\frac{1}{10} m/s

\frac{1}{15} m/s

15 m/s

10 m/s

answer is A.

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Detailed Solution

f = 20 cm = 0.2 m

$u = - 2 .8 m, v_{0} = 15 m/s$

$v_{I} = ?$

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

$\frac{1}{0.2} = \frac{1}{2.8} + \frac{1}{v}$

$\frac{1}{v} = \frac{1}{0.2} + \frac{1}{2.8}$

$\frac{1}{v} = \frac{3}{0.56}$

$v = \frac{56}{300} m$

$\frac{v_{I}}{v_{0}} = \frac{v^{2}}{u^{2}}$

$\frac{v_{I}}{15} = \frac{56 \times 56 \times 100}{300 \times 300 \times 28 \times 28}$

$v_{I} = \frac{1}{15} m/s$

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