Question

Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of $R=2m$. If the jogger is running at a speed of $5m{s}^{-1}$, how fast does the image of the jogger appeal to move when the jogger is $9m$ away?

• A. $\frac{1}{10}m{s}^{-1}$
• B. $\frac{1}{20}m{s}^{-1}$
• C. $\frac{1}{40}m{s}^{-1}$
• D. $\frac{1}{80}m{s}^{-1}$

Answer 1

Answer: (B)

$f=\frac{R}{2}=+\frac{2m}{2}=+1m$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f},u=-9m$
$\frac{1}{v}+\frac{1}{(-9)}=\frac{1}{(+1)}$
$\frac{1}{v}=\frac{1}{9}+\frac{1}{1}=\frac{10}{9}\text{ or }v=\frac{9}{10}m$
Now, differentiating the mirror equation w.r.t. time, we get,
$-\frac{1}{v^2}\frac{dv}{dt}-\frac{1}{u^2}\frac{du}{dt}=0$
$\Rightarrow \frac{dv}{dt}=-\left(\frac{v^2}{u^2}\right)\cdot \frac{du}{dt}={\left(\frac{1}{10}\right)}^2\cdot \left(5\frac{m}{s}\right)=\frac{5}{100}m{s}^{-1}$
$=\frac{1}{20}m{s}^{-1}$