一个质点沿直线运动，在经过定点 $O$ 后的 $t$ 秒时，它相对于 $O$ 的位移为 $s$ 米。质点的速度 $v\ \mathrm{ms}^{-1}$ 满足 $v = 6\cos 4t$。

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

When $t = 0$,
$v = 6\cos(4 \times 0)$
$v = 6\cos 0$
$v = 6(1) = 6$

1. Find the expression for acceleration by differentiating velocity

$a = \frac{dv}{dt}$
$a = \frac{d}{dt}(6\cos 4t) = 6(-4\sin 4t) = -24\sin 4t$

2. Set up the equation for acceleration and find the reference angle

$-24\sin 4t = 8$
$\sin 4t = -\frac{1}{3}$
$\text{Basic angle, } \alpha = \sin^{-1}\left(\frac{1}{3}\right) \approx 0.3398 \text{ rad}$

3. Solve for the first positive value of $t$

Since $\sin 4t$ is negative, $4t$ must be in the 3rd or 4th quadrant. The first positive value occurs in the 3rd quadrant:
$4t = \pi + 0.3398 \approx 3.4814$
$t = \frac{3.4814}{4} \approx 0.87035$
$t \approx 0.870 \text{ (to 3 s.f.)}$

1. Integrate the velocity function to find the general expression for displacement.

$s = \int v \, dt$
$s = \int 6\cos 4t \, dt$
$s = \frac{6}{4}\sin 4t + c$
$s = 1.5\sin 4t + c$

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

Since the particle passes through $O$ at $t = 0$, $s = 0$ when $t = 0$.
$0 = 1.5\sin(0) + c$
$c = 0$
$s = 1.5\sin 4t$

3. Substitute $t = 4$ to find the displacement.

When $t = 4$,
$s = 1.5\sin(4 \times 4)$
$s = 1.5\sin 16$
$s \approx -0.43185$
$s \approx -0.432\ \text{m (to 3 s.f.)}$

1. Find the expression for displacement $s$ in terms of $t$.

$s = \int v \, dt = \int 6\cos 4t \, dt$
$s = \frac{6}{4}\sin 4t + c = \frac{3}{2}\sin 4t + c$
Since $s = 0$ when $t = 0$,
$0 = \frac{3}{2}\sin(0) + c \implies c = 0$
$s = \frac{3}{2}\sin 4t$

2. Find the times when the particle is instantaneously at rest within the given interval.

Set $v = 0$:
$6\cos 4t = 0$
$\cos 4t = 0$
$4t = \frac{\pi}{2}, \frac{3\pi}{2}$
$t = \frac{\pi}{8}, \frac{3\pi}{8}$

3. Calculate the displacement at the critical times.

At $t = 0$, $s = 0$
At $t = \frac{\pi}{8}$, $s = \frac{3}{2}\sin\left(4 \times \frac{\pi}{8}\right) = \frac{3}{2}\sin\left(\frac{\pi}{2}\right) = \frac{3}{2}$
At $t = \frac{3\pi}{8}$, $s = \frac{3}{2}\sin\left(4 \times \frac{3\pi}{8}\right) = \frac{3}{2}\sin\left(\frac{3\pi}{2}\right) = -\frac{3}{2}$

4. Calculate the total distance travelled.

Distance from $t=0$ to $t=\frac{\pi}{8}$:
$d_1 = \left| \frac{3}{2} - 0 \right| = 1.5$
Distance from $t=\frac{\pi}{8}$ to $t=\frac{3\pi}{8}$:
$d_2 = \left| -\frac{3}{2} - \frac{3}{2} \right| = |-3| = 3$
Total distance $= 1.5 + 3 = 4.5\text{ m}$