Engineering
Physics
Physical Quantities
Question

A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8 m behind the first car is overtaking the first car at a 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 :

10 m/s

\(\frac{1}{{15}}{\mathop{\rm m}\nolimits} /s\)

\(\frac{1}{{10}}{\mathop{\rm m}\nolimits} /s\)

15 m/s

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Solution
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Mirror formula :

\(\frac{1}{v} + \frac{1}{{ - 280}} = \frac{1}{{20}}\)

\(\frac{1}{v} + \frac{1}{{20}} = \frac{1}{{280}}\)          

\(\frac{1}{v} + \frac{{14 + 1}}{{280}}\)

\(v = \frac{{280}}{{15}}\)

\({v_I} = - {\left( {\frac{v}{u}} \right)^2}.{v_{cm}}\)

\   \({v_I} = - {\left( {\frac{{280}}{{15 \times 280}}} \right)^2}.15\)

\   \({v_I} = \frac{{ - 15}}{{15 \times 15}}\)

\({v_I} = - \frac{1}{{15}}{\mathop{\rm m}\nolimits} /s\)

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