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A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate is the tip of his shadow moving? We've already set this up part of the way. We know that dx/dt=-2 meters per secon

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Answer:

It changes at a rate of 4/3 meter per second

Explanation:

In the given figure below we have

[tex]\Delta OBD\simeq \Delta ABC\\\\\therefore \frac{5}{X+Y}=\frac{2}{Y}\\\\[/tex]

Solving for Y given  [tex]X=2m/s[/tex] we get

[tex]\frac{5}{2+Y}=\frac{2}{Y}\\\\5Y=4+2Y\\\\Y=\frac{4}{3}m/s[/tex]

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