一个质点 $P$ 沿直线运动，在通过定点 $O$ 后的 $t$ 秒时，其速度 $v$ m/s 由 $v = 3t^2 + kt + 18$ 给出，其中 $k$ 为常数。当 $t = 1$ 时，质点的加速度为 $-9$ m/s$^2$。

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

$a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 + kt + 18) = 6t + k$

2. Substitute the given values of $t$ and $a$ to solve for $k$.

When $t = 1$, $a = -9$:
$6(1) + k = -9$
$k = -9 - 6$
$k = -15$ (shown)

1. Set velocity to zero and substitute the value of $k$

For the particle to be instantaneously at rest, $v = 0$.
Substitute $k = -15$ from part (i):
$3t^2 - 15t + 18 = 0$

2. Solve the quadratic equation for $t$

Divide by 3:
$t^2 - 5t + 6 = 0$
$(t - 2)(t - 3) = 0$
$t = 2 \text{ or } t = 3$

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

$s = \int v \, dt = \int (3t^2 - 15t + 18) \, dt$
$s = t^3 - \frac{15}{2}t^2 + 18t + c$
Since $s = 0$ when $t = 0$, $c = 0$.
$s = t^3 - \frac{15}{2}t^2 + 18t$

2. Calculate the displacement at the turning point $t = 2$

From part (ii), the particle is at rest at $t = 2$ and $t = 3$.
When $t = 2$,
$s = (2)^3 - \frac{15}{2}(2)^2 + 18(2)$
$s = 8 - 30 + 36 = 14$

3. Calculate the displacement at $t = 3$

When $t = 3$,
$s = (3)^3 - \frac{15}{2}(3)^2 + 18(3)$
$s = 27 - 67.5 + 54 = 13.5$

4. Calculate the total distance travelled

Distance travelled in first 2 seconds = $|14 - 0| = 14$ m
Distance travelled from $t = 2$ to $t = 3$ = $|13.5 - 14| = 0.5$ m
Total distance = $14 + 0.5 = 14.5$ m