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anonymous
 one year ago
As you stop your car at a stoplight, a rock becomes wedged between the tire treads. As you pull away from the light, the distance between the rock and the pavement varies sinusoidally with the distance you have traveled. The period is the circumference of the wheel. Assuming that the diameter of the wheel is 24 inches, find an equation that will model the height of the rock above the pavement.
A.
y = 12cos(1/6(x)) + 12
B.
y = 12cos(π/12(x)) + 12
C.
y = 12cos(π/6(x)) + 12
D.
y = 12cos(x/12) + 12
anonymous
 one year ago
As you stop your car at a stoplight, a rock becomes wedged between the tire treads. As you pull away from the light, the distance between the rock and the pavement varies sinusoidally with the distance you have traveled. The period is the circumference of the wheel. Assuming that the diameter of the wheel is 24 inches, find an equation that will model the height of the rock above the pavement. A. y = 12cos(1/6(x)) + 12 B. y = 12cos(π/12(x)) + 12 C. y = 12cos(π/6(x)) + 12 D. y = 12cos(x/12) + 12

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5if the period is given by the circumference of the wheel, then the angular frequency is: \[\Large \omega = \frac{{2\pi }}{T} = \frac{{2\pi }}{{2\pi R}} = \frac{1}{R} = \frac{1}{{12}}\] where R is the radius of the wheel

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5so, after a substitution, I get: \[\Large \begin{gathered} y\left( x \right) =  12\cos \left( {\omega x} \right) + 12 = \hfill \\ \hfill \\ =  12\cos \left( {\frac{x}{{12}}} \right) + 12 \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.5of course, in our case the period T is measured in meters and not in seconds
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