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.

Answer 1

Given that \(\frac{dx1}{dt}\) = speed of person = 60 cm/s 

Also \(\frac{dx2}{dt}\) = speed of tip of person's shadow 

Applying a similar triangle rule in Δ ABE & Δ DCE

\(\frac{4}{x_2} = \frac{1.6}{x_2-x_1}\)

2x₂-4x₁= 1.6x₂

2.4x₂ = 4x₁
differentiating on both sides w.r.t t

\(2.4\frac{dx_2}{dt}= 4\frac{dx_1}{dt}\)

= 100 cm/s

V_SP= V_SG - V_PG

V_{SP = 40 cm s ^-1}