Jamie is riding a Ferris wheel that takes fifteen seconds for each complete revolution. The diameter of the wheel is 10 meters and its cente

Question

Jamie is riding a Ferris wheel that takes fifteen seconds for each complete revolution. The diameter of the wheel is 10 meters and its center is 6 meters above the ground. (a) When Jamie is 9 meters above the ground and rising, at what rate (in meters per second) is Jamie gaining altitude? (b) When is Jamie rising most rapidly? At what rate?

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Allison 1 month 2021-10-19T01:33:24+00:00 1 Answer 0 views 0

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    2021-10-19T01:35:06+00:00

    Answer:

    The answers to the question is

    (a) Jamie is gaining altitude at 1.676 m/s

    (b) Jamie rising most rapidly at t = 15 s

    At a rate of 2.094 m/s.

    Step-by-step explanation:

    (a) The time to make one complete revolution = period T = 15 seconds

    Here will be required to develop the periodic motion equation thus

    One complete revolution = 2π,

    therefore the  we have T = 2π/k = 15

    Therefore k = 2π/15

    The diameter = radius of the wheel = (diameter of wheel)/2 = 5

    also we note that the center of the wheel is 6 m above ground

    We write our equation in the form

    y = 5*sin(\frac{2*\pi*t}{15} )+6

    When Jamie is 9 meters above the ground and rising we have

    9 = 5*sin(\frac{2*\pi*t}{15} )+6 or 3/5 = sin(\frac{2*\pi*t}{15} ) = 0.6

    which gives sin⁻¹(0.6) = 0.643 =\frac{2*\pi*t}{15}

    from where t = 1.536 s

    Therefore Jamie is gaining altitude at

    \frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) = 1.676 m/s.

    (b) Jamie is rising most rapidly when   the velocity curve is at the highest point, that is where the slope is zero

    Therefore we differentiate the equation for the velocity again to get

    \frac{d^2y}{dx^2} = -5*(\frac{\pi *2}{15} )^2*sin(\frac{2\pi t}{15}) =0, π, 2π

    Therefore -sin(\frac{2\pi t}{15} ) = 0 whereby t = 0 or

    \frac{2\pi t}{15} = π and t =  7.5 s, at 2·π t = 15 s

    Plugging the value of t into the velocity equation we have

    \frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi t}{15}) = – 2/3π m/s which is decreasing

    so we try at t = 15 s and we have \frac{dy}{dt} = 5*\frac{\pi *2}{15} *cos(\frac{2\pi *15}{15}) = \frac{2}{3} \pim/s

    Hence Jamie is rising most rapidly at t = 15 s

    The maximum rate of Jamie’s rise is 2/3π m/s or 2.094 m/s.

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