A person of height 1 . 6m is walking away from a lamp post of...

Given:

Speed of person $= \dfrac{d{x}_{1}}{dt} = 60cm{s}^{- 1}$

Also Speed of tip of person's shadow $= \dfrac{d{x}_{2}}{dt}$.

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

$\dfrac{4}{{x}_{2}} = \dfrac{1 . 6}{{x}_{2} - {x}_{1}}$

$⇒ 4{x}_{2} - 4{x}_{1} = 1 . 6{x}_{2}$

$⇒ 2 . 4{x}_{2} = 4{x}_{1}$

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

$2 . 4\dfrac{d{x}_{2}}{dt} = 4\dfrac{d{x}_{1}}{dt}$

$⇒ \dfrac{d{x}_{2}}{dt} = \dfrac{4}{2 . 4}\left(60\right)$

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

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

$⇒ {v}_{SP} = 100cm{s}^{- 1} - 60cm{s}^{- 1}$

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