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.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.