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29 October, 12:46

A Ferris wheel is 40 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 10 minutes. How many minutes of the ride are spent higher than 34 meters above the ground?

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  1. 29 October, 16:38
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    t = 2.9517 min

    Step-by-step explanation:

    Given

    D = 40 m ⇒ R = D/2 = 40 m/2 = 20 m

    ybottom = 2 m

    ytop = ybottom + D = 40 m + 2 m = 42 m

    yref = 34 m

    t = 10 min

    The height above the ground (y) is a sinusoidal function.

    The minimum height is ybottom = 2 m

    The maximum height is ytop = 42 m;

    The midline is (ybottom + ytop) / 2 = (2 m + 42 m) / 2 = 22 m

    If we model the wheel as follows

    x² + y² = R²

    where

    y = yref - (R + ybottom) = 34 m - (20 m + 2 m) = 12 m

    R = 20 m

    we have

    x² + (12 m) ² = (20) ²

    ⇒ x = 16 m

    then

    tan (θ/2) = x/y

    ⇒ tan (θ/2) = 16 m/12 m

    ⇒ θ = 106.26°

    Knowing the angle of the circular sector, we apply the relation

    t = (106.26°) * (10 min/360°)

    ⇒ t = 2.9517 min

    Since the period of revolution is 10 minutes, the ride is above 34 meters for 2.9517 minutes each revolution.
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