A car is fitted with a convex side-view mirro | vedclass

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(B) Given: Focal length $f = +20\ cm = +0.2\ m$,Object distance $u = -2.8\ m$,Speed of object $\frac{du}{dt} = -15\ m/s$ (since the distance is decrea

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$\frac{1}{15}\ m/s$

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(B) Given: Focal length $f = +20\ cm = +0.2\ m$,Object distance $u = -2.8\ m$,Speed of object $\frac{du}{dt} = -15\ m/s$ (since the distance is decreasing).Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Differentiating with respect to time $t$: $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$.So,$\frac{dv}{dt} = -\frac{v^2}{u^2} \frac{du}{dt}$.From the mirror formula,$v = \frac{uf}{u-f} = \frac{(-2.8)(0.2)}{-2.8 - 0.2} = \frac{-0.56}{-3.0} = \frac{0.56}{3} = \frac{56}{300} = \frac{14}{75}\ m$.Now,$\frac{dv}{dt} = -\left( \frac{v}{u} \right)^2 \frac{du}{dt} = -\left( \frac{14/75}{-2.8} \right)^2 (-15) = -\left( \frac{14/75}{-210/75} \right)^2 (-15) = -\left( -\frac{14}{210} \right)^2 (-15) = -\left( -\frac{1}{15} \right)^2 (-15) = -\left( \frac{1}{225} \right) (-15) = \frac{15}{225} = \frac{1}{15}\ m/s$.

$(i)$ 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	$(d) \; 22 \; m/s$

$A$ 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

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