A person of height 1.6 m is walking away from a lamp post of height 4 m along a straight path on the flat ground. The lamp post and the person are always perpendicular to the ground. If the speed of the person is 60 cm , the speed of the tip of the person

Question

A person of height 1.6 m is walking away from a lamp post of height 4 m along a straight path on the flat ground. The lamp post and the person are always perpendicular to the ground. If the speed of the person is 60 cm $s^{- 1}$ , the speed of the tip of the person's shadow on the ground with respect to the person is _____ cm $s^{- 1}$ . [2023]

Smart Achievers · Accepted Answer

(40)

$From figure, ∆ A B E and ∆ C D E are similar triangles, so$

$\frac{4}{y} = \frac{1.6}{y - x}$ $\Rightarrow 4 (y - x) = 1.6 y$

$\Rightarrow 4 y - 4 x = 1.6 y$

$\Rightarrow 2.4 y = 4 x$

$∴$ $x = 0.6 y$

Differentiating both sides,

$\frac{d x}{d t} = 0.6 \frac{d y}{d t}$

$(\frac{d x}{d t} = speed of person = 60 cm/s and \frac{d y}{d t} = speed of tip of person's shadow)$

$60 = 0.6 \frac{d y}{d t}$ $∴$ $\frac{d y}{d t} = 100 cm/s$

Speed of tip of person's shadow on the ground, w.r.t. person $= 100 - 60 = 40 cm/s$

Solution

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