A Ferris wheel...

Question

A Ferris wheel is 35 meters in diameter and boarded from a platform that is 5 meters above the ground. The six o'clock position on the Ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 6 minutes. How many minutes of the ride are spent higher than 30 meters above the ground?

Answer 1

We are interested in the height (in meters) of the riders as a function of time (in minutes) so h(t) = (some function of time). This, of course is a cyclical function (sine wave type function) but to determine the details, we need to think about what is given. The rider gets on (at t = 0) at the minimum. To me, this is closest matched to a negative cosine function.

So, we'll start with h(t) = -Acos[B(t - C)] + D where:

A is the amplitude (the up/down from the midline) which is 35/2 or 17.5

B is determined by the period where B = 2π/period = 2π/6 = π/3

C is the phase shift, so, in this case is zero

D is the midline which is 5 meters plus 17.5 = 22.5

So h(t) = -17.5cos[(π/3)(t - 0)] + 22.5 or

h(t) = -17.5cos[(π/3)t] + 22.5

To find the amount of time above 30 meters, we let h = 30:

30 = -17.5cos[(π/3)t] + 22.5

7.5 = -17.5cos[(π/3)t]

-0.42857 = cos[(π/3)t]

cos^-1(-0.42857) = (π/3)t

2.0137 = (π/3)t

t = 1.923 minutes

At an angle of 2.0137 is the 1st place the cosine is -0.42857 (which is in Q2). Here is a sketch of it in the Unit Circle (shown in red):

The other angle where cosine is -0.42857 is in Q3 (shown in blue) and can be attained by subtracting 2.0137 from 2π which gives us:

4.269478 = (π/3)t

t = 4.077 minutes

So the rider is above 30 meters between t = 1.923 and t = 4.077 or for 2.154 minutes