A Ferris wheel has a radius of 15 meters and the bottom of the Ferris wheel is 2

Question

A Ferris wheel has a radius of 15 meters and the bottom of the Ferris wheel is 2.2 meters above the ground. Elliot boards the Ferris wheel at the 3 o'clock position. The Ferris wheel rotates at a constant speed of s radians per second.

Write a formula that gives the angle measure, a, (in radians) swept out from the 3 o'clock position in terms of the number of seconds elapsed, t, since Elliot boarded the Ferris wheel.

Write a formula that gives the period, b, (in seconds) in terms of the speed of the Ferris Wheel, s, (in radians per second).

Write a formula that gives Elliot's height above the ground, h, (in meters), in terms of the number of seconds elapsed, t, since Elliot boarded the Ferris wheel.

Explanation / Answer

Radius = 15
2.2 m above ground

So, min = 2.2
max = 2.2 + 15 = 17.2

Amplitude = (max - min)/2
A = 15/2
A = 7.5

Midline = (max + min)/2
D = (2.2 + 17.2)/2
D = 9.7

C = 0 because it starts at midline, which is true for sin.

B = 2pi/period

Speed = s rad/sec
PEriod = 2pi/s

And thus B = 2pi/(2pi / s)
B = s

So, we have
Y = Asin[B(x - c)] + D

y = 7.5sin(st) + 9.7 ----> ANSWER for part 3

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Period in terms of speed...

We had period = 2pi/s ----> ANSWER for part 2

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Angle measure :
s rad/sec

So, in t seconds,
we cover s*t radians

So, s*t ---> ANS for part 1

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