下图显示了海潮的垂直位移 $y$ m，由 $y = 2 - \frac{3}{2} \cos pt$ 表示，其中 $t$ 是以小时为单位的时间（自最低位移处开始计算），$p$ 是一个常数。垂直位移再次达到最低点需要 12 个小时。

1. Identify the period of the function from the given information

The problem states that it takes 12 hours for the vertical displacement to be at its lowest point again. This means the period of the function is $12$ hours.

2. Relate the period to the constant $p$ in the equation

For a function $y = a \cos(pt) + c$, the period is given by $\frac{2\pi}{p}$.
Therefore, $\frac{2\pi}{p} = 12$.

3. Solve for $p$

$12p = 2\pi$
$p = \frac{2\pi}{12}$
$p = \frac{\pi}{6}$ (shown)

1. Substitute the given values into the equation and isolate the trigonometric term

Substitute $y = 1.6$ and $p = \frac{\pi}{6}$ into $y = 2 - \frac{3}{2} \cos pt$:
$1.6 = 2 - \frac{3}{2} \cos\left(\frac{\pi}{6}t\right)$
$\frac{3}{2} \cos\left(\frac{\pi}{6}t\right) = 2 - 1.6$
$\frac{3}{2} \cos\left(\frac{\pi}{6}t\right) = 0.4$
$\cos\left(\frac{\pi}{6}t\right) = \frac{0.4}{1.5} = \frac{4}{15}$

2. Find the basic angle and the first two possible values for the argument

Basic angle $\alpha = \cos^{-1}\left(\frac{4}{15}\right) \approx 1.3011$ rad
Since $\cos\left(\frac{\pi}{6}t\right)$ is positive, the angle is in the 1st and 4th quadrants:
$\frac{\pi}{6}t = 1.3011 \quad \text{or} \quad \frac{\pi}{6}t = 2\pi - 1.3011$
$\frac{\pi}{6}t = 1.3011 \quad \text{or} \quad \frac{\pi}{6}t = 4.9820$

3. Solve for $t$

$t = \frac{1.3011 \times 6}{\pi} \quad \text{or} \quad t = \frac{4.9820 \times 6}{\pi}$
$t \approx 2.485 \quad \text{or} \quad t \approx 9.515$
$t = 2.48 \quad \text{or} \quad t = 9.52 \text{ (to 3 s.f.)}$

1. Solve for the first value of $t$ from part (ii) to establish the safe time interval

From part (ii), $2 - \frac{3}{2} \cos\left(\frac{\pi}{6}t\right) = 1.6$
$\cos\left(\frac{\pi}{6}t\right) = \frac{4}{15}$
$\frac{\pi}{6}t = 1.3013 \implies t \approx 2.485$
The first time the displacement reaches 1.6 m is at $t = 2.49$ hours.

2. Convert the given clock times to $t$ values

Given $t=0$ is 08 00, the time 09 00 corresponds to $t = 1$ and 10 00 corresponds to $t = 2$.

3. Explain the safety based on the interval

Since the displacement starts at 0.5 m (at $t=0$) and only reaches 1.6 m at $t = 2.49$, the displacement $y$ is strictly less than 1.6 m for $0 \leqslant t < 2.49$.
Because the interval $t=1$ to $t=2$ falls entirely within this safe period, it is safe to swim between 09 00 and 10 00.