Question

A car is fitted with a convex side-view mirror 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. $\frac{1}{10}m/s$
• B. $\frac{1}{15}m/s$
• C. $10m/s$
• D. $15m/s$

Answer 1

Answer: (B)

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

The velocity of the first car

$\frac{du}{dt}=15\text{m}{\text{s}}^{-1}=1500\text{c}{\text{m}}^{-1}$

The focal length of the mirror $\left(f\right)=+20\text{cm}$

The object distance is $\left(u\right)=-2.8\text{m}=-280\text{cm}$

With the help of the mirror formula, we get-

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\frac{1}{v}=\frac{1}{20}-\frac{1}{-280}$

$\frac{1}{v}=\frac{1}{20}+\frac{1}{280}$

$\frac{1}{v}=\frac{15}{280}$

We differentiate the mirror formula with respect to the time and get-

$0=-\frac{1}{v^2}.\frac{dv}{dt}-\frac{1}{u^2}.\frac{du}{dt}$

$\frac{1}{v^2}.\frac{dv}{dt}=-\frac{1}{u^2}.\frac{du}{dt}$

${\left(\frac{15}{280}\right)}^2.\frac{dv}{dt}=-{\left(\frac{1}{-280}\right)}^2.\left(1500\right)$

$\frac{225}{{\left(280\right)}^2}.\frac{dv}{dt}=-\frac{1}{{\left(280\right)}^2}.\left(1500\right)$

$\frac{dv}{dt}=\frac{1500}{225}\text{cm}{\text{s}}^{-1}$

$\frac{dv}{dt}=\frac{1500}{22500}\text{m}{\text{s}}^{-1}$

$\frac{dv}{dt}=\frac{1}{15}\text{m}{\text{s}}^{-1}$

So, the correct option is option (B).