A man of height 2 m walks at a uniform speed | vedclass

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Question

(A) Let $AB$ be the lamp post of height $6 \ m$ and $CD$ be the man of height $2 \ m$ at a particular time $t$.Let $AC = x$ be the distance of the man

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(A) Let $AB$ be the lamp post of height $6 \ m$ and $CD$ be the man of height $2 \ m$ at a particular time $t$.Let $AC = x$ be the distance of the man from the lamp post and $CE = y$ be the length of his shadow.Given that the man walks at a speed of $5 \ km/hr$,so $\frac{dx}{dt} = 5 \ km/hr$.We need to find the rate at which the shadow increases,i.e.,$\frac{dy}{dt}$.Since $	riangle ABE \sim 	riangle CDE$,we have:$\frac{AB}{CD} = \frac{AE}{CE}$$\frac{6}{2} = \frac{x + y}{y}$$3 = \frac{x + y}{y}$$3y = x + y$$2y = x$Differentiating both sides with respect to $t$:$2 \frac{dy}{dt} = \frac{dx}{dt}$$2 \frac{dy}{dt} = 5$$\frac{dy}{dt} = \frac{5}{2} = 2.5 \ km/hr$.Thus,the length of the shadow increases at the rate of $2.5 \ km/hr$.

$A$ man of height $2 \ m$ walks at a uniform speed of $5 \ km/hr$ away from a lamp post which is $6 \ m$ high. The rate at which the length of his shadow increases is .......... $km/hr$.

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