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Motion

Section: Physics

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Question : 2 of 12

Marks: +1, -0

A person of height

1.6 \text{m}

is walking away from a lamp post of height

4 \text{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 \text{cm} \text{s}^{-1}

, the speed of the tip of the person's shadow on the ground with respect to the person is __________

\text{cm} \text{s}^{-1}

[JEE Adv 2023 P1]

Your Answer:

Solution:

Given that

\frac{d x_1}{d t} =

speed of person

=60 \text{cm} / \text{s}

Also

\; \frac{d x_2}{d t} =

speed of tip of person's shadow

Applying similar triangle rule in

\triangle A B E \& \triangle D C E

\; \frac{4}{x_2} = \; \frac{1.6}{x_2 - x_1}

4 x_2 - 4 x_1 = 1.6 x_2

2.4 x_2 = 4 x_1

Differentiate both sides w.r.t.

t

2.4 \; \frac{d x_2}{d t} = 4 \; \frac{d x_1}{d t}

\; \frac{d x_2}{d t} \; = \; \frac{4}{2.4} (60)

\; = 100 \text{cm} / \text{s}

\vec{v}_{SP} \; = \vec{v}_{SG} - \vec{v}_{PG}

v_{SP} \; = 100 \text{cm} \text{s}^{-1} - 60 \text{cm} \text{s}^{-1}

\; = 40 \text{cm} \text{s}^{-1}

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