Dean L. answered • 01/25/21

BS, MST in Math; 10+ years tutoring/teaching Algebra 2

The Ferris Wheel radius is 40 feet and the center of the wheel is 50 feet high, lowest point of the wheel is at a height of 50 – 40 = 10 feet and the highest point at 50 + 40 = 90 feet.

Since there are 12 hours on a clockface each hour represents 1/12 of a full revolution so each ‘hour’ is rotated through in 45/12 = 3.75 seconds.

Therefore the time to rotate to t o’clock from 3 o’clock , t < 12, is (t – 3)(3.75) seconds.

For Part 1 we have:

a. (10- 3)(3.75) = (7)(3.75) = 26.25 sec

b. (7- 3)(3.75) = (4)(3.75) = 15 sec

c. (4- 3)(3.75) = (1)(3.75) = 3.75 sec

For Part 2, the beginning platform height is at 3 o’clock corresponding to 0^{o} and a height 50 feet.

Because the speed of revolution is 45 seconds, the ratio of the time t seconds after passing 3 o'clock position and 45 seconds gives the proportional part of the 360^{o} the platform has turned through, namely an angle of (t/45)( 360^{o}) = (8t)^{o}. Then the vertical distance y feet from the platform to a horizontal line through the center at 0^{o} is 40sin(8t). Recall that lowest point of the wheel is at a height of 10 feet and the center at 0^{o} and is at 50 feet. Therefore, the platform height is 50 – y = 50 – sin(8t) feet.

What is the platform height off the ground at each of these times?

a. 3 seconds after passing 3 o'clock position. h = 50 – sin(8*3) = 50 – sin(24) = 33.73 feet

b. 12 seconds after passing the 3 o'clock position. h = 50 – sin(8*12) = 50 – sin(96) = 10.22 feet

c. 24 seconds after passing the 3 o'clock position. h = 50 – sin(8*24) = 50 – sin(192) = 53.32 feet

d. 35 seconds after passing the 3 o'clock position. h = 50 – sin(8*35) = 50 – sin(280) = 89.39 feet