QJEEAdv23Phae56: JEE Advanced 2023 Physics - Class 11 JEEAdv23Phae56 Solution: A person of height is walking away from a lamp post of height…

Given that $\frac{d x_1}{d t}=$ speed of person $=60 \mathrm{~cm} / \mathrm{s}$



Also $\frac{d x_2}{d t}=$ speed of tip of person's shadow

Applying similar triangle rule in $\triangle A B E$ and $\triangle D C E$

$\begin{aligned}& \frac{4}{x_2}=\frac{1.6}{x_2-x_1} \\\\& 4 x_2-4 x_1=1.6 x_2 \\\\& 2.4 x_2=4 x_1\end{aligned}$

Differentiate both sides w.r.t. $t$

$\begin{aligned}& 2.4 \frac{d x_2}{d t}=4 \frac{d x_1}{d t} \\\\& \frac{d x_2}{d t}=\frac{4}{2.4}(60) \\\\&=100 \mathrm{~cm} / \mathrm{s}\end{aligned}$

This is the speed of the tip of the person's shadow with respect to the lamp post. But, we need to find the speed of the shadow's tip with respect to the person, which is the relative speed :

$\begin{aligned}\vec{v}_{S P} & =\vec{v}_{S G}-\vec{v}_{P G} \\\\v_{S P} & =100 \mathrm{~cm} \mathrm{~s}^{-1}-60 \mathrm{~cm} \mathrm{~s}^{-1} \\\\& =40 \mathrm{~cm} \mathrm{~s}^{-1}\end{aligned}$