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 0o 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 360o the platform has turned through, namely an angle of (t/45)( 360o) = (8t)o. Then the vertical distance y feet from the platform to a horizontal line through the center at 0o is 40sin(8t). Recall that lowest point of the wheel is at a height of 10 feet and the center at 0o 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