Question

Figure shows a concave mirror of focal length $10\text{cm}$. A point object on the principal axis at a distance of $15\text{cm}$ from the pole starts moving away from the pole and its position-time relation is given by $x=-(5t^2+10t+15)\text{cm/s}$. The speed of the image at $t=1\text{s}$ will be

• A. $6\text{cm/s}$
• B. $7\text{cm/s}$
• C. $5\text{cm/s}$
• D. $4\text{cm/s}$

Answer 1

The correct option is C $5\text{cm/s}$

Given:
Object distance, $x=-(5t^2+10t+15)$
focal length, $f=-10\text{cm}$

From mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}-\frac{1}{(5t^2+10t+15)}=-\frac{1}{10}$

$\Rightarrow \frac{1}{v}=\frac{-1}{10}+\frac{1}{5t^2+10t+15}$

$\Rightarrow \frac{1}{v}=\frac{-5t^2-10t-15+10}{10(5t^2+10t+15)}$

$\Rightarrow \frac{1}{v}=\frac{-5t^2-10t-5}{10(5t^2+10t+15)}$

$\Rightarrow \frac{1}{v}=\frac{-t^2-2t-1}{2(5t^2+10t+15)}$

$\Rightarrow v=\frac{10t^2+20t+30}{-t^2-2t-1}$

Differentiating both sides w.r.t $t$

$\Rightarrow \frac{dv}{dt}=\frac{(-t^2-2t-1)(20t+20)-(10t^2+20t+30)(-2t-2)}{{(-t^2-2t-1)}^2}$

$\Rightarrow \frac{dv}{dt}=\frac{(10t^2+20t+30)(2t+2)-(20t+20)(t^2+2t+1)}{(t^2+2t+1{)}^2}$

$\Rightarrow \frac{dv}{dt}=\frac{40t+40}{{(t^2+2t+1)}^2}$

At $t=1\text{s}$,

$\frac{dv}{dt}=\frac{40+40}{{(1+2+1)}^2}=\frac{80}{16}=5\text{cm/s}$

Hence, $\left(c\right)$ is the correct option.