Ferris Wheel Evacuation Drill
The Ferris wheel on a pier has a diameter of 18 m. The floor of each gondola is 2 m above the deck at the lowest point of the ride, and the wheel makes one full turn every 40 s at constant speed. Riders board at the lowest point, so a rider's height in metres after t seconds is h(t) = -9cos(πt/20) + 11. The operator's insurance policy says that during an evacuation drill no gondola may sit above 15 m for more than 15 s in a single revolution, and the operator must also report, for a 10-minute ride, the total time a gondola spends above 15 m. The wheel normally runs for exactly 10 minutes, which is 15 complete turns.
1. Find every time in the first revolution (0 <= t <= 40) at which a gondola is exactly 15 m above the deck.
Read the worked answer
Set the model equal to 15:
-9cos(πt/20) + 11 = 15 -9cos(πt/20) = 4 cos(πt/20) = -4/9 ≈ -0.4444
The reference angle is arccos(4/9) ≈ 1.110 rad (about 63.6°). Since the cosine is negative, the solutions for the angle θ = πt/20 lie in quadrants II and III:
θ₁ = π - 1.110 ≈ 2.031 rad θ₂ = π + 1.110 ≈ 4.252 rad
Convert with t = 20θ/π (factor ≈ 6.366): t₁ ≈ 12.9 s and t₂ ≈ 27.1 s
Sanity check: the maximum height occurs at t = 20 s, and the two answers are symmetric about it (each about 7.1 s away), which is what a cosine model requires. Heights above 15 m occur between these two times, centred on the top of the ride.
2. Does one revolution satisfy the insurance rule? Report the total time above 15 m for the full 10-minute ride.
Read the worked answer
The gondola is above 15 m between the two crossing times:
duration = (θ₂ - θ₁)(20/π) = (4.252 - 2.031)(6.366) ≈ 2.221 × 6.366 ≈ 14.1 s
That is less than 15 s, so one revolution satisfies the policy, but only by about 0.9 s. There is no room for a slower boarding pause at the top; the operator should note the narrow margin.
For the full ride, 600 s at 40 s per turn is exactly 15 complete revolutions. Each revolution contributes the same 14.1 s, so the reported total is:
15 × 14.1 ≈ 212 s, or about 3 min 32 s
Because the ride is a whole number of revolutions, there is no partial-turn correction to make.
3. A new motor makes one revolution take 36 s instead of 40 s. Does the rule still hold, and what happens to the reported 10-minute total?
Read the worked answer
The new model is h(t) = -9cos(2πt/36) + 11 = -9cos(πt/18) + 11. The amplitude and vertical shift are unchanged, so the cosine value at the crossings is still -4/9; only the angle-to-time conversion changes.
The fraction of each revolution spent above 15 m depends only on the angle interval, not on the period: fraction = (4.252 - 2.031)/(2π) ≈ 2.221/6.283 ≈ 0.3534
New duration per revolution = 0.3534 × 36 ≈ 12.7 s, comfortably under 15 s, so the rule is still met.
For the 10-minute report, note that 600 s is now 16.67 revolutions, not a whole number. The correct method is to apply the fraction to the whole ride rather than multiply a single-revolution time by a whole number of turns: 0.3534 × 600 ≈ 212 s. The total is unchanged, because speeding up the wheel reduces the time per revolution but gives proportionally more revolutions.