Question 22 - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at May 29, 2023 by Teachoo

A man 1.6 m tall walks at the rate of 0.3 m/sec away from a street light that is 4 m above the ground. At what rate is the tip of his shadow moving? At what rate is his shadow lengthening?

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Question 22 A man 1.6 m tall walks at the rate of 0.3 m/sec away from a street light that is 4 m above the ground. At what rate is the tip of his shadow moving? At what rate is his shadow lengthening?
Let AB be the lamp post
& MN be the man of height 2m.
& AM = x meter
& MS is the shadow of the man
Let length of shadow MS = s meter
Given man walks at speed of 0.3 m/sec
∴ 𝒅𝒙/𝒅𝒕 = 0.3 m/s
We need to find rate at which length of his shadow is increasing i.e. we need to find 𝒅𝒔/𝒅𝒕
In ΔASB
tan θ = 𝐴𝐵/𝐴𝑆
tan θ =𝟒/(𝒙 + 𝒔)
In ∆ MSN
tan θ = 𝑀𝑁/𝑀𝑆
tan θ =(𝟏.𝟔)/𝒔
From (1) & (2)
𝟒/(𝒙 + 𝒔) = (𝟏.𝟔)/𝒔
4s = 1.6x + 1.6s
4s – 1.6s = 1.6x
2.4s = 1.6x
24/10 𝑠=16/10 𝑥
24s = 16x
3s = 2x
2x = 3s
We need to find 𝑑𝑠/𝑑𝑡
Now,
2x = 3s
Diff w.r.t t
(𝑑(2𝑥))/𝑑𝑡= 𝑑(3𝑠)/𝑑𝑡
2𝒅𝒙/𝒅𝒕= 3.𝑑𝑠/𝑑𝑡
2 × 0.3 = 3 𝑑𝑠/𝑑𝑡
𝑑𝑠/𝑑𝑡 = (2 × 0.3)/3
𝒅𝒔/𝒅𝒕 =𝟎.𝟐 m/s
So, 𝒅𝒔/𝒅𝒕 = 𝟓/𝟐 km/hr.

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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