Question

A man, $2\, m$ 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)

$A B=$ street light, $C=$ Man.
From $\triangle A B E$ and $\triangle D C E$,
$\frac{A B}{C D}=\frac{A E}{C E}=\frac{A C+C E}{C E}$
$\Rightarrow \frac{A B}{C D}=\frac{A C}{C E}+1$
$\Rightarrow \frac{A C}{C E}=\frac{A B}{C D}-1$
$\Rightarrow \frac{A C}{C E}=\frac{5 \frac{1}{3}}{2}-1=\frac{5}{3}$
$\Rightarrow C E=\frac{3}{5} A C$
Differentiating with respect to time we get,
$\frac{d}{d t} C E=\frac{3}{5}\left(\frac{d}{d t} A C\right) \ldots$.(i)
None $\frac{d}{d t} A C=-1 \frac{2}{3} m s^{-1}$
$\frac{d}{d t} A C=-\frac{5}{3} m s^{-1}$
Negative sign as $A C$ decreases with time. substituting in Eq. (i) we get.
Rate of change of shadow length
$=\frac{3}{5}\left(-\frac{5}{3}\right)=-1\, m s^{-1}$