JEE MAIN - Physics (2011 - No. 3)
A car is fitted with a convex side-view mirror of focal length $$20$$ $$cm$$. A second car $$2.8m$$ 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 :
$${1 \over {15}}\,m/s$$
$$10\,m/s$$
$$15\,m/s$$
$${1 \over {10}}\,m/s$$
Explanation
From mirror formula
$${1 \over v} + {1 \over u} = {1 \over f}\,\,\,$$
so, $$\,\,\,{{dv} \over {dt}} = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$$
$$ \Rightarrow {{dv} \over {dt}} = - {\left( {{f \over {u - f}}} \right)^2}{{du} \over {dt}}$$
$$ \Rightarrow {{dv} \over {dt}} = {1 \over {15}}m/s$$
$${1 \over v} + {1 \over u} = {1 \over f}\,\,\,$$
so, $$\,\,\,{{dv} \over {dt}} = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$$
$$ \Rightarrow {{dv} \over {dt}} = - {\left( {{f \over {u - f}}} \right)^2}{{du} \over {dt}}$$
$$ \Rightarrow {{dv} \over {dt}} = {1 \over {15}}m/s$$
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