JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 3)
Car B overtakes another car A at a relative speed of 40 ms$$-$$1. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A?
4 ms$$-$$1
0.2 ms$$-$$1
40 ms$$-$$1
0.1 ms$$-$$1
Explanation
Here, $${1 \over v} + {1 \over u} = {1 \over f}$$ .......(1)
$$ \Rightarrow $$ $${1 \over v} + {1 \over { - 190}} = {1 \over {10}}$$
$$ \Rightarrow $$ v = $${{19} \over 2}$$
Differentiating equation (1) w.r.t t we get
$$ - {1 \over {{v^2}}}\left( {{{dv} \over {dt}}} \right) - {1 \over {{u^2}}}\left( {{{du} \over {dt}}} \right) = 0$$
$$ \Rightarrow $$ $$\left( {{{dv} \over {dt}}} \right) = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$$
$$ \Rightarrow $$ $$\left( {{{dv} \over {dt}}} \right) = - {\left( {{{{{19} \over 2}} \over {190}}} \right)^2} \times 40$$
$$ = {1 \over {400}} \times 40 = 0.1$$ m/s
$$ \Rightarrow $$ $${1 \over v} + {1 \over { - 190}} = {1 \over {10}}$$
$$ \Rightarrow $$ v = $${{19} \over 2}$$
Differentiating equation (1) w.r.t t we get
$$ - {1 \over {{v^2}}}\left( {{{dv} \over {dt}}} \right) - {1 \over {{u^2}}}\left( {{{du} \over {dt}}} \right) = 0$$
$$ \Rightarrow $$ $$\left( {{{dv} \over {dt}}} \right) = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)$$
$$ \Rightarrow $$ $$\left( {{{dv} \over {dt}}} \right) = - {\left( {{{{{19} \over 2}} \over {190}}} \right)^2} \times 40$$
$$ = {1 \over {400}} \times 40 = 0.1$$ m/s
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