Ferris wheel revolution

Question

A Ferris wheel is 50 meters in diameter and boarded from a platform that is 3 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. 

The function H(t)=−25cos(π30t)+28 gives your height in meters above the ground t seconds after the wheel begins to turn. 

it takes 60 seconds for 1 revolution

2. How many seconds of the ride (1 full revolution) are spent higher than 29 meters above the ground? 

Answer 1

I thought the equation was H(t)=−25cos(π30t)+28 not cos(pi/30 *t), right?

Answer 2

Let's see at what time the height equals 29 meters. If we set 28 - 25cos(πt/30) = 29, we can solve for t by doing some algebra and using the inverse cosine function.

28 - 25cos(πt/30) = 29

-25cos(πt/30) = 1

cos(πt/30) = -1/25

πt/30 = arccos(-1/25)

t = (30/π) * arccos(-1/25) ≈ 15.3820738 seconds.

So we get one answer for when h(t) = 29 meters. But we also know that if we went around the wheel in reverse starting from t = 60 seconds, then it would take the same amount of time to get up to 29 meters as it did going around the forward direction. Thus we can also say that when t ≈ 60 - 15.3820738 = 44.6179262 seconds, the height will also be at 29 meters.

So finally, the length of this time interval is Δt ≈ 44.6179262 - 15.3820738 = 29.2358524 seconds during any single revolution that the ride is above 29 meters.

Answer 3

-25cos(π/30t) + 28 > 29; cos(π/30t) < -1/25; cos^-1(-1/25) < π/30t < 2π - cos^-1(-1/25);

1.611 <π/30t < 4.672; 15.384 < t < 44.614. Time in secunds