Question

A man, $2m$ tall, walks at the rate of $1\frac{2}{3}m/s$ towards a street light which is $5\frac{1}{3}m$ above the ground. The rate at which the length of his shadow is changing when he is $3\frac{1}{3}m$ m away from the base of the light is________

• A. -1 m / s
• B. 2 m / s
• C. -2 m / s
• D. 1 m / s

Answer 1

Answer: (A)

$AB=$ street light, $C=$ Man.
From $△ABE$ and $△DCE$,
$\frac{AB}{CD}=\frac{AE}{CE}=\frac{AC+CE}{CE}$
$\Rightarrow \frac{AB}{CD}=\frac{AC}{CE}+1$
$\Rightarrow \frac{AC}{CE}=\frac{AB}{CD}-1$
$\Rightarrow \frac{AC}{CE}=\frac{5\frac{1}{3}}{2}-1=\frac{5}{3}$
$\Rightarrow CE=\frac{3}{5}AC$
Differentiating with respect to time we get,
$\frac{d}{dt}CE=\frac{3}{5}\left(\frac{d}{dt}AC\right)\dots$.(i)
None $\frac{d}{dt}AC=-1\frac{2}{3}m{s}^{-1}$
$\frac{d}{dt}AC=-\frac{5}{3}m{s}^{-1}$
Negative sign as $AC$ decreases with time. substituting in Eq. (i) we get.
Rate of change of shadow length
$=\frac{3}{5}\left(-\frac{5}{3}\right)=-1m{s}^{-1}$