一座半径为 15 m 的摩天轮每 3 分钟旋转一周。乘客在摩天轮的最低点上车，该点距离地面 1 m。假设乘客在上车点登车后摩天轮开始旋转。乘客登车 $t$ 分钟后距离地面的高度 $h$ m 可以用如下方程表示：$h = - a \cos b t + c.$

1. Determine the value of $a$ (amplitude).

The radius of the ferris wheel is 15 m, which corresponds to the amplitude of the cosine function.
$a = 15$

2. Determine the value of $c$ (vertical shift).

The lowest point is 1 m above the ground. The minimum value of $h = -a \cos bt + c$ occurs when $\cos bt = 1$, giving $h = -a + c$.
$-15 + c = 1$
$c = 16$

3. Determine the value of $b$ (related to period).

The ferris wheel completes one rotation every 3 minutes, so the period is 3.
Period $= \frac{2\pi}{b} = 3$
$b = \frac{2\pi}{3}$

1. Recall the equation for $h$ from part (a) and differentiate it with respect to $t$ to find the velocity $v$.

From part (a), $h = -15 \cos\left(\frac{2\pi}{3}t\right) + 16$.

$v = \frac{dh}{dt} = \frac{d}{dt}\left[-15 \cos\left(\frac{2\pi}{3}t\right) + 16\right]$
$v = -15 \left[-\sin\left(\frac{2\pi}{3}t\right)\right] \left(\frac{2\pi}{3}\right)$
$v = 10\pi \sin\left(\frac{2\pi}{3}t\right)$

2. Deduce the maximum speed from the velocity expression.

The maximum value of $\sin\left(\frac{2\pi}{3}t\right)$ is $1$.
Therefore, the maximum speed is $|v|_{\max} = 10\pi(1) = 10\pi$ m/min.

1. Differentiate the velocity expression to find the acceleration function

From part (b), $v = 10\pi \sin\left(\frac{2\pi}{3}t\right)$.
$a = \frac{dv}{dt} = 10\pi \left(\frac{2\pi}{3}\right) \cos\left(\frac{2\pi}{3}t\right)$
$a = \frac{20\pi^2}{3} \cos\left(\frac{2\pi}{3}t\right)$

2. Substitute $t = 1$ and find the magnitude of the acceleration

When $t = 1$,
$a = \frac{20\pi^2}{3} \cos\left(\frac{2\pi}{3}\right)$
$a = \frac{20\pi^2}{3} \left(-\frac{1}{2}\right) = -\frac{10\pi^2}{3}$
Magnitude of acceleration $= \left|-\frac{10\pi^2}{3}\right| = \frac{10\pi^2}{3} \approx 32.9 \text{ m/min}^2$ (to 3 s.f.)