A Ferris wheel has a radius of 11 meters and the bottom of t

A Ferris wheel has a radius of 11 meters and the bottom of the Ferris wheel is 1 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.

a. 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.

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

c. 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.

Solution

a) constant speed of s radians per second

angle measure , a = s*t (radians)

b) Period = time for one revolution p = 2pi/s

c) h(t) = 12 -11sin(st)

at st=0; h(0) = 12 --->3 \'o clock position

at st = 90 deg ; h(t) = 12 -11 = 1mt

at st = 180 deg ; h(t) = 12 -0 = 12 mt

at st= 270 deg h(t) = 12+11 = 23 mt

So, h(t) =  12 -11sin(st)