Question

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

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