JEE Advance - Physics (2023 - Paper 1 Online - No. 12)
A person of height $1.6 \mathrm{~m}$ is walking away from a lamp post of height $4 \mathrm{~m}$ along a straight path on the flat ground. The lamp post and the person are always perpendicular to the ground. If the speed of the person is $60 \mathrm{~cm} \mathrm{~s}^{-1}$, the speed of the tip of the person's shadow on the ground with respect to the person is ____________ $\mathrm{cm}~ \mathrm{s}^{-1}$.
Answer
40
Explanation
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 :
$$ \begin{aligned} \vec{v}_{S P} & =\vec{v}_{S G}-\vec{v}_{P G} \\\\ v_{S P} & =100 \mathrm{~cm} \mathrm{~s}^{-1}-60 \mathrm{~cm} \mathrm{~s}^{-1} \\\\ & =40 \mathrm{~cm} \mathrm{~s}^{-1} \end{aligned} $$
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 :
$$ \begin{aligned} \vec{v}_{S P} & =\vec{v}_{S G}-\vec{v}_{P G} \\\\ v_{S P} & =100 \mathrm{~cm} \mathrm{~s}^{-1}-60 \mathrm{~cm} \mathrm{~s}^{-1} \\\\ & =40 \mathrm{~cm} \mathrm{~s}^{-1} \end{aligned} $$
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