Question

Car $B$ overtakes another car $A$ at a relative speed of $40{\text{ms}}^{-1}$. How fast will the image of car $B$ appear to move, in the mirror of focal length $10\text{cm}$, fitted in car $A$, when the car $B$ is $1.9\text{m}$ away from the car $A$?

• A. $4{\text{ms}}^{-1}$
• B. $0.1{\text{ms}}^{-1}$
• C. $40{\text{ms}}^{-1}$
• D. $0.2{\text{ms}}^{-1}$

Answer 1

The correct option is B $0.1{\text{ms}}^{-1}$

$V_{B,A}=10{\text{ms}}^{-1}$
$f=10\text{cm}=0.1\text{m}$
$u=1.9\text{m}$

As the focal length is positive, it is a concave mirror.
Now, $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\frac{1}{-0.1}=\frac{1}{v}+\frac{1}{-1.9}$

$\frac{1}{v}=\frac{1}{1.9}-\frac{1}{0.1}=\frac{18}{1.9}$

$\therefore v=-0.1055\text{m}$

From the mirror equation, we can write,
$\Rightarrow f^{-1}=u^{-1}+v^{-1}$

Differentiating both sides w.r.t. time,

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

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

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

Here, $\frac{du}{dt}\to \text{speed of object w.r.t. mirror}$
$\frac{dv}{dt}\to \text{speed of image w.r.t. mirror}$

Now, $V_{B,A}=\frac{du}{dt}=40{\text{ms}}^{-1}$
$u=-1.9\text{m},v=-0.1055\approx -0.1\text{m}$

$\therefore \frac{1}{(-1.9{)}^2}\times 40=-\frac{1}{(-0.1{)}^2}\times \frac{dv}{dt}$

After calculating, we get, $\frac{dv}{dt}=-0.11{\text{m/s}}^{-1}$

Negative sign indicated that, as object is moving towards the mirror, the image is moving away from it.

Hence, $\left(A\right)$ is the correct answer.