A point object is moving on the principal axi | vedclass

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(C) For a concave mirror,the mirror formula is $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Differentiating with respect to time $t$,we get $-\frac{1}{f^

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$4 \; cm/s$ away from the mirror

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(C) For a concave mirror,the mirror formula is $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Differentiating with respect to time $t$,we get $-\frac{1}{f^2} \frac{df}{dt} = -\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt}$.Since $f$ is constant,$\frac{df}{dt} = 0$,so $\frac{dv}{dt} = -(\frac{v}{u})^2 \frac{du}{dt}$.Here,$v_i = \frac{dv}{dt}$ and $v_o = \frac{du}{dt}$.Given $f = -24 \; cm$,$u = -60 \; cm$,and $v_o = -9 \; cm/s$ (moving towards the mirror).Using $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} = \frac{1}{-24} - \frac{1}{-60} = \frac{-5 + 2}{120} = \frac{-3}{120} = -\frac{1}{40}$,so $v = -40 \; cm$.The velocity of the image is $v_i = -(\frac{v}{u})^2 v_o = -(\frac{-40}{-60})^2 (-9) = -(\frac{2}{3})^2 (-9) = -(\frac{4}{9})(-9) = 4 \; cm/s$.The positive sign indicates the image is moving away from the mirror.

$(i)$ Velocity of image	$(a) \; 2 \; m/s$
$(ii)$ Velocity of image with respect to mirror	$(b) \; 20 \; m/s$
$(iii)$ Velocity of image with respect to object	$(c) \; 11 \; m/s$
$(iv)$ Velocity of image if mirror is stopped	$(d) \; 22 \; m/s$

$A$ point object is moving on the principal axis of a concave mirror of focal length $24 \; cm$ towards the mirror. When it is at a distance of $60 \; cm$ from the mirror,its velocity is $9 \; cm/s$. What is the velocity of the image at that instant?

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