Question

A car is fitted with a convex side-view mirror of focal length $20cm$. A second car $2.8m$ behind the first car is overtaking the first car at a relative speed of $15m/s$. The speed of the image of the second car as seen in the mirror of the first one is :

• A. $10m/s$
• B. $15m/s$
• C. $\frac{1}{10}m/s$
• D. $\frac{1}{15}m/s$

Answer 1

The correct option is D $\frac{1}{15}m/s$

REF.Image

$\frac{du}{dt}={\vartheta }_0=15m{s}^{-1}$

We have,

focal length of the mirror (f)= +20 cm

object distance 'u' = -280 cm

and object velocity $\left(\frac{du}{dt}\right)=15m{s}^{-1}$

We know that,

$\frac{1}{\vartheta }+\frac{1}{u}=\frac{1}{f}$ ___ (i) (mirror formula)

Where, $\vartheta \to$ image distance.

$\frac{1}{\vartheta }=\frac{1}{280}=\frac{1}{20}$

$\Rightarrow \frac{1}{\vartheta }=\frac{1}{20}+\frac{1}{280}=\frac{15}{280}$

$\Rightarrow \frac{1}{\vartheta }=\frac{15}{280}$

differentiating (i) with respect to time

$\frac{-1}{{\vartheta }^2}\frac{d\vartheta }{dt}-\frac{1}{u^2}\frac{du}{dt}=0$

$\Rightarrow -\frac{15}{280}\times \frac{15}{280}{\vartheta }_i=\frac{1}{280}\times \frac{1}{280}\times 15$

$\Rightarrow {\vartheta }_i=\frac{du}{dt}=\frac{1}{15}m{s}^{-1}$