Question

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

• A. $\frac{1}{15}\text{m/s}$
• B. $\frac{1}{20}\text{m/s}$
• C. $\frac{1}{25}\text{m/s}$
• D. $\frac{1}{30}\text{m/s}$

Answer 1

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

For the given convex mirror,
$u=-2.8\text{m}=-280\text{cm},f=+20\text{cm}$
Using mirror formula,
$\frac{1}{v}-\frac{1}{280}=\frac{1}{20}$
$\Rightarrow \frac{1}{v}=\frac{3}{56}$
$\Rightarrow v=\frac{56}{3}\text{cm}$
So, $m=\frac{-v}{u}=\frac{-\frac{56}{3}}{-280}=\frac{1}{15}$
Also, velocity of the image when the object is moving along the principal axis,
$v_{i,m}=-m^2{v}_{o,m}$
$\Rightarrow v_{i,m}=-{\left(\frac{1}{15}\right)}^2\times 15=-\frac{1}{15}\text{m/s}$
Hence, the speed of the image of the second car as seen in the mirror of the first car will be : $\frac{1}{15}\text{m/s}$.
$\begin{array}{c}\text{Why this question?}\\ \text{Tips: As relative velocity is given, thus we can}\\ \text{assume mirror of first car to be in rest and can apply}\\ \text{formula with sign convention.}\end{array}$