JEE Advanced 2023 — Paper 1 — PHY-III solution

JEE Advanced2023Paper 1PHY-III

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 _______ $cm s^{- 1}$ .

Your answer

Solution

Similar triangles. Let $x$ be the person's distance from the lamp post and $y$ the distance of the shadow tip from the lamp post. Similar triangles give $\frac{4}{y} = \frac{1.6}{y - x}$ , which simplifies to $4 (y - x) = 1.6 y$ , or $2.4 y = 4 x$ , i.e. $x = 0.6 y$ .

Speed of shadow tip (ground frame). Differentiating: $\frac{d x}{d t} = 0.6 \frac{d y}{d t}$ . With $\frac{d x}{d t} = 60 cm s^{- 1}$ : $\frac{d y}{d t} = 100 cm s^{- 1}$ .

Relative to the person. $\frac{d y}{d t} - \frac{d x}{d t} = 100 - 60 = 40 cm s^{- 1}$ .

Answer: $40 cm s^{- 1}$ .

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