一质点在直线上运动，使得在离开固定点 $O$ 之后 $t$ 秒，其速度 $v$ 由 $v = \frac{36}{(t-3)^2} - 9$ 给出。求

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

$v = 36(t-3)^{-2} - 9$
$a = \frac{dv}{dt} = 36(-2)(t-3)^{-3} = -\frac{72}{(t-3)^3}$

2. Substitute $t = 0$ to find the initial acceleration

When $t = 0$,
$a = -\frac{72}{(0-3)^3} = -\frac{72}{-27} = \frac{8}{3}$

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

$v = 0$
$\frac{36}{(t-3)^2} - 9 = 0$
$\frac{36}{(t-3)^2} = 9$

2. Solve the equation for $t$

$(t-3)^2 = \frac{36}{9}$
$(t-3)^2 = 4$
$t - 3 = \pm 2$
$t = 3 + 2 \text{ or } t = 3 - 2$
$t = 5 \text{ or } t = 1$

1. Identify the time when the particle is at instantaneous rest within the interval $0 \leqslant t \leqslant 2$.

From part (ii), $v = 0$ when $t = 1$ and $t = 5$.
Within the first 2 seconds, the particle changes direction at $t = 1$.

2. Find the indefinite integral of velocity to get the displacement function.

Let $s(t)$ be the displacement.
$s(t) = \int \left( \frac{36}{(t-3)^2} - 9 \right) dt$
$s(t) = \int \left( 36(t-3)^{-2} - 9 \right) dt$
$s(t) = \frac{36(t-3)^{-1}}{-1} - 9t + C$
$s(t) = -\frac{36}{t-3} - 9t + C$

3. Evaluate the displacement at $t=0$, $t=1$, and $t=2$.

Since we only need changes in displacement, we can ignore $C$ (or set $C=0$ for the definite integral evaluation):
At $t = 0$, value $= -\frac{36}{-3} - 0 = 12$
At $t = 1$, value $= -\frac{36}{-2} - 9 = 18 - 9 = 9$
At $t = 2$, value $= -\frac{36}{-1} - 18 = 36 - 18 = 18$

4. Calculate the total distance travelled by summing the absolute displacements of each interval.

Distance from $t=0$ to $t=1$:
$|9 - 12| = |-3| = 3$
Distance from $t=1$ to $t=2$:
$|18 - 9| = |9| = 9$
Total distance $= 3 + 9 = 12$