A person of height 1 . 6   m is walking away from a lamp post of height 4   m along a straight…

Given:

Speed of person $=\frac{d{x}_1}{dt}=60 \mathrm{cm} {\mathrm{s}}^{-1}$

Also Speed of tip of person's shadow $=\frac{d{x}_2}{dt}$. 

As $∆ABE$ and $∆DCE$ are similar triangles, therefore we can write

$\frac{4}{x_2}=\frac{1.6}{x_2-x_1}$

$\Rightarrow 4x_2-4x_1=1.6x_2$

$\Rightarrow 2.4x_2=4x_1$

Differentiate both sides w.r.t. $t$, we get

$2.4\frac{d{x}_2}{dt}=4\frac{d{x}_1}{dt}$

$\Rightarrow \frac{d{x}_2}{dt}=\frac{4}{2.4}\left(60\right)$

$=100 \mathrm{cm} {\mathrm{s}}^{-1}$

Now, speed of the tip of the person's shadow on the ground with respect to the person will be ${\overrightarrow{v}}_{SP}={\overrightarrow{v}}_{SG}-{\overrightarrow{v}}_{PG}$

$\Rightarrow v_{SP}=100 \mathrm{cm} {\mathrm{s}}^{-1}-60 \mathrm{cm} {\mathrm{s}}^{-1}$

$=40 \mathrm{cm} {\mathrm{s}}^{-1}$