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

VSP= VSG - VPG

VSP = 40 cm s -1