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

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(A) Let $AB$ be the lamp-post and $PQ$ be the man. Let $C$ be the tip of the shadow. Let $AP = x$ be the distance of the man from the lamp-post and $P

Vedclass · Accepted Answer

$2$

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(A) Let $AB$ be the lamp-post and $PQ$ be the man. Let $C$ be the tip of the shadow. Let $AP = x$ be the distance of the man from the lamp-post and $PC = y$ be the length of his shadow.Given: $AB = 9 \ m$,$PQ = 2 \ m$,and $\frac{dx}{dt} = 7 \ m/min$.Since $	riangle CAB$ and $	riangle CPQ$ are similar triangles,we have:$\frac{PC}{AC} = \frac{PQ}{AB}$$\frac{y}{x+y} = \frac{2}{9}$$9y = 2x + 2y$$7y = 2x$$x = \frac{7}{2}y$Differentiating both sides with respect to time $t$:$\frac{dx}{dt} = \frac{7}{2} \frac{dy}{dt}$Substituting $\frac{dx}{dt} = 7$:$7 = \frac{7}{2} \frac{dy}{dt}$$\frac{dy}{dt} = 2 \ m/min$.Thus,the length of his shadow increases at a rate of $2 \ m/min$.

$A$ man of height $2 \ m$ walks at a uniform speed of $7 \ m/min$ away from a lamp post of height $9 \ m$. The rate (in $m/min$) at which the length of his shadow increases is

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