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Rina8888 [55]
3 years ago
9

A 16 foot ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 3 feet per second, how fa

st is the bottom of the ladder moving along the ground when the bottom of the ladder is 4 feet from the wall? Leave your answer in exact form, simplified as much as possible. Do not include units.
Physics
1 answer:
JulsSmile [24]3 years ago
7 0

Answer:

11.625

Explanation:

L = length of the ladder = 16 ft

v_{y} = rate at which top of ladder slides down = - 3 ft/s

v_{x} = rate at which bottom of ladder slides

y = distance of the top of ladder from the ground

x = distance of bottom of ladder from wall = 4 ft

Using Pythagorean theorem

L² = x² + y²

16² = 4² + y²

y = 15.5 ft

Also using Pythagorean theorem

L² = x² + y²

Taking derivative both side relative to "t"

0 = 2x\frac{\mathrm{d} x}{\mathrm{d} t} + 2y\frac{\mathrm{d} y}{\mathrm{d} t}

0 = x v_{x} + y v_{y}

0 = 4 v_{x} + (15.5) (- 3)

v_{x} = 11.625 ft/s

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Answer:

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We are given that

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\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s

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r=-\frac{15}{2\pi}t+\frac{168}{2\pi}

E=-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}

E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}

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E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})

E=103.1 V

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