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

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Question

(D) The mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Differentiating with respect to time $t$,we get $-\frac{1}{v^2} \frac{dv}{dt} - \f

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Differentiating with respect to time $t$,we get $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$.Thus,the velocity of the image $v_I = \frac{dv}{dt} = -(\frac{v}{u})^2 \frac{du}{dt}$.Since $\frac{v}{u} = \frac{f}{u-f}$,we have $v_I = -(\frac{f}{u-f})^2 v_O$.Given $f = 20\ cm = 0.2\ m$,$u = -2.8\ m$,and $v_O = -15\ m/s$ (approaching the mirror).Substituting the values: $v_I = -(\frac{0.2}{-2.8 - 0.2})^2 	imes (-15) = -(\frac{0.2}{-3.0})^2 	imes (-15) = -(\frac{1}{15})^2 	imes (-15) = \frac{1}{15}\ m/s$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

$A$ 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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