一个质点 P 沿直线运动，在时间为 $t$ 秒时，其速度 $v$ m/s 由下式给出： $v = \cos\left(\dfrac{\pi}{4}t + \dfrac{\pi}{6}\right)$。 P 相对于固定点 O 的初始位移为 $\dfrac{1}{\pi}$ m。

1. Substitute $t = 0$ into the velocity equation to find the initial velocity

When $t = 0$,
$v = \cos\left(\frac{\pi}{4}(0) + \frac{\pi}{6}\right)$
$v = \cos\left(\frac{\pi}{6}\right)$

2. Evaluate the exact value of the trigonometric function

$v = \frac{\sqrt{3}}{2}$

1. Set velocity to zero for instantaneous rest

$v = \cos\left(\frac{\pi}{4}t + \frac{\pi}{6}\right) = 0$

2. Find the first two positive angles where cosine is zero

$\frac{\pi}{4}t + \frac{\pi}{6} = \frac{\pi}{2}, \frac{3\pi}{2}$

3. Solve for the corresponding values of t

For the first value:
$\frac{\pi}{4}t = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$
$t = \frac{4}{3}$

For the second value:
$\frac{\pi}{4}t = \frac{3\pi}{2} - \frac{\pi}{6} = \frac{4\pi}{3}$
$t = \frac{16}{3}$

1. Find the expression for displacement $s$ by integrating velocity $v$.

$s = \int \cos\left(\frac{\pi}{4}t + \frac{\pi}{6}\right) dt$
$s = \frac{4}{\pi} \sin\left(\frac{\pi}{4}t + \frac{\pi}{6}\right) + c$

2. Use the initial condition to find the constant of integration $c$.

When $t = 0$, $s = \frac{1}{\pi}$.
$\frac{1}{\pi} = \frac{4}{\pi} \sin\left(\frac{\pi}{6}\right) + c$
$\frac{1}{\pi} = \frac{4}{\pi} \left(\frac{1}{2}\right) + c$
$\frac{1}{\pi} = \frac{2}{\pi} + c$
$c = -\frac{1}{\pi}$
So, $s = \frac{4}{\pi} \sin\left(\frac{\pi}{4}t + \frac{\pi}{6}\right) - \frac{1}{\pi}$

3. Identify the times when the particle changes direction in the interval $0 \le t \le 4$.

From part (b), $v = 0$ when $\frac{\pi}{4}t + \frac{\pi}{6} = \frac{\pi}{2} \implies \frac{\pi}{4}t = \frac{\pi}{3} \implies t = \frac{4}{3}$.
The next time is when $\frac{\pi}{4}t + \frac{\pi}{6} = \frac{3\pi}{2} \implies \frac{\pi}{4}t = \frac{4\pi}{3} \implies t = \frac{16}{3}$, which is outside the interval $0 \le t \le 4$.
So the particle changes direction only at $t = \frac{4}{3}$ in the first 4 seconds.

4. Calculate the displacement at $t=0$, $t=\frac{4}{3}$, and $t=4$, and find the total distance.

At $t = 0$, $s_0 = \frac{1}{\pi}$.
At $t = \frac{4}{3}$, $s_{4/3} = \frac{4}{\pi} \sin\left(\frac{\pi}{2}\right) - \frac{1}{\pi} = \frac{4}{\pi}(1) - \frac{1}{\pi} = \frac{3}{\pi}$.
At $t = 4$, $s_4 = \frac{4}{\pi} \sin\left(\pi + \frac{\pi}{6}\right) - \frac{1}{\pi} = \frac{4}{\pi} \left(-\frac{1}{2}\right) - \frac{1}{\pi} = -\frac{2}{\pi} - \frac{1}{\pi} = -\frac{3}{\pi}$.

Distance travelled = $|s_{4/3} - s_0| + |s_4 - s_{4/3}|$
$= \left|\frac{3}{\pi} - \frac{1}{\pi}\right| + \left|-\frac{3}{\pi} - \frac{3}{\pi}\right|$
$= \frac{2}{\pi} + \frac{6}{\pi} = \frac{8}{\pi}$ m.