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Question

A Ferris wheel is boarded from a platform that is 5 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. 

The function H(t)=−20cos(π38t)+25 gives your height in meters above the ground t seconds after the wheel begins to turn.How many seconds of the ride (1 full revolution) are spent higher than 33 meters above the ground?

diameter is 40 meters, and takes 76 seconds for 1 revolution

Answer 1

H(t)=−20cos(π38t) + 25 > 33; cos(πt/38) < - 0.4 ; cos^-1(- 0.4) < πt/38 < 2π - cos^-1(- 0.4);

1.9823 <πt/38 < 4.3009; 23.9775 < t < 52.0227; Δt = 52.0227 - 23.9775 = 28.045 s

Answer 2

Draw a circle of radius 20 to represent the ferris wheel.

Draw a horizontal line 8 units above the center that intersects the circle at points A and B,

Draw the triangle with vertices at A, B, and the center. Draw the altitude from the center to the horizontal line. The altitude has length 8.

Let θ be the interior angle of the triangle with vertex at the center of the circle.

cos( θ/2) = 8/20 = 0.4. So, θ = 2.3186 radians

Length of smaller arc AB = (20ft)(2.3186) = 46.37 ft

Let t = time that the ferris wheel is above 33 ft

t / 76 = 46.37 ft / Circumference

t / 76 = 46.37 / (40π)

So, t = 28.04 sec

Answer 3

33 < -20 cos (38πt) + 25

Can you solve for t and answer?