A street light is mounted at the top of a 15ft tall pole. A man 6ft tall walks away from the pole with a speed of 5ft/sec along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?

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A street light is mounted at the top of a 15ft tall pole. A man 6ft tall walks away from the pole with a speed of 5ft/sec along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?

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Answer 1 · 2018-01-17T09:45:14

$8.33 \frac{f t .}{sec .}$ $8.33(ft.)/(sec.)$

Explanation:

The street light is mounted at the top of a $15 f t$ $15ft$ tall pole. Let us consider the man $6 f t$ $6ft$ tall $x f t$ $xft$ away from the pole. His shadow forms two ends - one end is at his feet and the shadow extends away from the pole till the tip of the shadow.

Let this be depicted by the figure shown below.
enter image source here

Here the distance of the man from lamp post be $x f t .$ $xft.$ and let his shadow be $y f t$ $yft$ from man. Now as man is moving away from lamp post, $x$ $x$ is a function of $t$ $t$ and speed of man is $\frac{d x}{d t}$ $(dx)/(dt)$

Then, as they form similar triangle, we have

$\frac{15}{15 - 6} = \frac{x + y}{x}$ $15/(15-6)=(x+y)/x$ i.e. $15 x = 9 x + 9 y$ $15x=9x+9y$ or $9 y = 6 x$ $9y=6x$ and $y = \frac{2}{3} x$ $y=2/3x$

and shadow is $x + \frac{2}{3} x = \frac{5}{3} x$ $x+2/3x=5/3x$ from lamp post. And hence when man moves $δ x$ $deltax$ feet, shadow moves $\frac{5}{3} δ x$ $5/3deltax$ feet

and hence, shadow moves with a speed of $\frac{5}{3} \frac{d x}{d t}$ $5/3(dx)/(dt)$ i.e. $\frac{5}{3} \times 5 = \frac{25}{3} = 8.33 \frac{f t .}{sec .}$ $5/3xx5=25/3=8.33(ft.)/(sec.)$

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A street light is mounted at the top of a 15ft tall pole. A man 6ft tall walks away from the pole with a speed of 5ft/sec along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?

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