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A laser pointer is placed on a platform that rotates at a rate of 60 revolutions per minute. The beam hits a wall 10 m away, producing a dot of light that moves horizontally along the wall. Let θθ be the angle between the beam and the line through the searchlight perpendicular to the wall.

Respuesta :

Answer:

[tex]v = 20\pi sec^2\theta[/tex]

Explanation:

As we know that the angle made with the vertical is given so we have

[tex]tan\theta = \frac{x}{10}[/tex]

so we have

[tex]x = 10 tan\theta[/tex]

now differentiate both sides with time

[tex]v = \frac{dx}{dt} = 10 sec^2\theta \frac{d\theta}{dt}[/tex]

[tex]v = 10 \omega sec^2\theta[/tex]

now we know that

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi (\frac{60}{60})[/tex]

[tex]\omega = 2\pi[/tex]

now we have

[tex]v = 10(2\pi) sec^2\theta[/tex]

[tex]v = 20\pi sec^2\theta[/tex]

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