一质点从点 $O$ 由静止出发并沿直线运动，其速度 $v$ m/s 由 $v = 24t - 6t^2$ 给出，其中 $t$ 是其运动开始后的时间（以秒为单位）。

1. Set velocity to zero to find when the particle is instantaneously at rest

$v = 0$
$24t - 6t^2 = 0$

2. Solve the quadratic equation for $t$

$6t(4 - t) = 0$
$t = 0 \text{ or } t = 4$
Since the particle starts from rest at $t = 0$, it is instantaneously at rest again at $t = 4$.

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

$s = \int v \, dt = \int (24t - 6t^2) \, dt$
$s = 12t^2 - 2t^3 + c$

2. Determine the constant of integration using initial conditions.

Since the particle starts from $O$, $s = 0$ when $t = 0$.
$0 = 12(0)^2 - 2(0)^3 + c \implies c = 0$
So, $s = 12t^2 - 2t^3$.

3. Find the time when the particle returns to the starting point ($s = 0$).

Set $s = 0$:
$12t^2 - 2t^3 = 0$
$2t^2(6 - t) = 0$
$t = 0$ (start) or $t = 6$

1. Find the expression for acceleration by differentiating velocity with respect to time.

$a = \frac{dv}{dt}$
$a = \frac{d}{dt}(24t - 6t^2)$
$a = 24 - 12t$

2. Evaluate the acceleration at $t = 2$.

When $t = 2$,
$a = 24 - 12(2)$
$a = 0$

3. State the conclusion based on the calculated acceleration.

Since $a = 0$, the particle is not accelerating (it is moving at a constant velocity at that instant).

1. Identify the time when the particle changes direction

From part (a), the particle is instantaneously at rest when $v = 0$, which occurs at $t = 4$. Since $v > 0$ for $0 < t < 4$ and $v < 0$ for $t > 4$, the particle changes direction at $t = 4$.

2. Find the displacement function $s(t)$

$s = \int v \, dt = \int (24t - 6t^2) \, dt$
$s = 12t^2 - 2t^3 + c$
Since the particle starts from $O$, $s = 0$ when $t = 0$, so $c = 0$.
$s = 12t^2 - 2t^3$

3. Calculate the displacement at the key times $t=4$ and $t=7$

At $t = 4$:
$s(4) = 12(4)^2 - 2(4)^3 = 192 - 128 = 64$
At $t = 7$:
$s(7) = 12(7)^2 - 2(7)^3 = 588 - 686 = -98$

4. Calculate the total distance travelled

Distance travelled from $t = 0$ to $t = 4$ is $|64 - 0| = 64\text{ m}$.
Distance travelled from $t = 4$ to $t = 7$ is $|-98 - 64| = |-162| = 162\text{ m}$.
Total distance $= 64 + 162 = 226\text{ m}$.