Question

It is night. Someone who is 4 feet tall is walking away from a street light at a rate of 8 feet per second. The street light is 12 feet tall. The person casts a shadow on the ground in front of them. How fast is the length of the shadow growing when the person is 3 feet from the street light?

Answer 1

Answer: 4 ft/s

Explanation:

Given

height of man$=4 ft$

speed of person $v=8 ft/s$

height if street light$=12 ft$

Let x be the distance between person and street light and y be the length of his shadow

From diagram

as the two triangle ADE and ABC are similar therefore we can say that

$\frac{4}{12}=\frac{y}{x+y}$

$\frac{1}{3}=\frac{y}{x+y}$

$x+y=3y$

$x=2y$

differentiate above Equation w.r.t time we get

$\frac{\mathrm{d} x}{\mathrm{d} t}=2\frac{\mathrm{d} y}{\mathrm{d} t}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{8}{2}=4 ft/s$