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

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(B) Let $y$ be the distance of the man from the lamp post and $x$ be the length of his shadow.Given that the man walks away from the lamp post at a sp

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$\frac{5}{2} \, m/h$

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(B) Let $y$ be the distance of the man from the lamp post and $x$ be the length of his shadow.Given that the man walks away from the lamp post at a speed of $5 \, m/h$,we have $\frac{dy}{dt} = 5 \, m/h$.From the similar triangles in the figure,we have the relationship:$\frac{x}{2} = \frac{x + y}{6}$Multiplying both sides by $6$,we get:$3x = x + y$$2x = y$$x = \frac{1}{2}y$Differentiating both sides with respect to time $t$,we get:$\frac{dx}{dt} = \frac{1}{2} \frac{dy}{dt}$Substituting $\frac{dy}{dt} = 5 \, m/h$,we get:$\frac{dx}{dt} = \frac{1}{2} 	imes 5 = \frac{5}{2} \, m/h$.Thus,the rate at which the length of his shadow increases is $\frac{5}{2} \, m/h$.

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

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