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3 years ago

On your drive home from school, you travel 1200 m in 80 s, then 400 min 20 s.

Physics

1 answer:

Mama L [17]3 years ago

5 0

Average speed = Total distance/total time taken

$= \frac{1200 + 400}{80 + 20}$

$= \frac{1600}{100}$

$= \frac{16}{1}$

$= 16 \: ms {}^{ - 1}$

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A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

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Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

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Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

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The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

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$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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