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 \(\frac{dx_1}{dt}\) = speed of person = 60 cm/s

Also \(\frac{dx_2}{dt}\) = speed of tip of person's shadow

Applying similar triangle rule in \(\Delta\)ABE & \(\Delta\)DCE