The speed of the tip of her shadow would be 5.714 ft/sec. ft/sec.
let H, be the height of the pole = 20 ft.
h, be the height of the woman = 6 ft.
x, be the length of the woman's shadow,
and s, be the distance from the pole to the woman = 15
by using the equation of similar triangles,
[tex]\frac{H}{h} = \frac{(s+x)}{x}[/tex] or Hx = h(x+s)
differentiating, [tex]H\frac{dx}{dt} = h (\frac{ds}{dt} + \frac{dx}{dt} )[/tex]
[tex]20\frac{dx}{dt} = 6(4 + \frac{dx}{dt} ) = 24 + 6 \frac{dx}{dt}[/tex]
[tex]14\frac{dx}{dt} = 24[/tex],
[tex]\frac{dx}{dt} = \frac{24}{14}[/tex] ≈ 1.714 ft/sec, is how fast her shadow is lengthening.
the speed of the tip of her shadow would be the speed of shadow lengthening + speed of her traveling = 1.714 + 4 = 5.714 ft/sec.
thus, a street light is at the top of a 20 ft tall pole. a woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. the speed of the tip of her shadow would be 5.714 ft/sec.
know more about the speed of the shadow problems here: https://brainly.com/question/13316193
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