一个质点沿直线运动，在离开固定点 $O$ 后 $t$ 秒时，其速度 $v\text{ m s}^{-1}$ 由 $v = t^2 - 2kt - 6k$ 给出，其中 $k$ 为常数。
质点的最小速度发生在 $t = 2$ 时。

1. Differentiate the velocity function with respect to $t$ to find the acceleration.

$v = t^2 - 2kt - 6k$
$\frac{dv}{dt} = 2t - 2k$

2. Use the condition for minimum velocity to solve for $k$.

For minimum velocity, $\frac{dv}{dt} = 0$.
Given this occurs when $t = 2$:
$2(2) - 2k = 0$
$4 - 2k = 0$
$2k = 4$
$k = 2 \text{ (shown)}$

1. Substitute $k=2$ into the velocity equation and integrate to find displacement $s$.

Given $k = 2$, $v = t^2 - 4t - 12$.
$s = \int v \, dt$
$s = \int (t^2 - 4t - 12) \, dt$
$s = \frac{t^3}{3} - 2t^2 - 12t + C$

2. Find the constant of integration using the initial condition.

When $t = 0$, the particle is at $O$, so $s = 0$.
$0 = \frac{0^3}{3} - 2(0)^2 - 12(0) + C \implies C = 0$
$s = \frac{1}{3}t^3 - 2t^2 - 12t$

3. Set $s = 0$ to find when the particle returns to $O$.

$\frac{1}{3}t^3 - 2t^2 - 12t = 0$
$t(t^2 - 6t - 36) = 0$
$t = 0$ (at start) or $t^2 - 6t - 36 = 0$

4. Solve the quadratic equation to show there is only one positive root.

For $t^2 - 6t - 36 = 0$:
$t = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-36)}}{2(1)}$
$t = \frac{6 \pm \sqrt{36 + 144}}{2}$
$t = \frac{6 \pm \sqrt{180}}{2} = 3 \pm 3\sqrt{5}$
Since $t \ge 0$, $t = 3 + 3\sqrt{5}$ (approx $9.71$ s).
The other root $t = 3 - 3\sqrt{5} < 0$ is rejected.
Since there is only one valid positive value for $t$, the particle returns to $O$ only once.

1. Find the time when the particle is instantaneously at rest

From (i), $k = 2$, so $v = t^2 - 4t - 12$.
Set $v = 0$ to find when the particle changes direction:
$t^2 - 4t - 12 = 0$
$(t - 6)(t + 2) = 0$
Since $t \ge 0$, $t = 6$.

2. Calculate the displacement at $t = 0$, $t = 6$, and $t = 7$

From (ii), the displacement is $s = \frac{t^3}{3} - 2t^2 - 12t$.
At $t = 0$, $s = 0$.
At $t = 6$, $s = \frac{6^3}{3} - 2(6)^2 - 12(6) = 72 - 72 - 72 = -72\text{ m}$.
At $t = 7$, $s = \frac{7^3}{3} - 2(7)^2 - 12(7) = \frac{343}{3} - 98 - 84 = -67\frac{2}{3}\text{ m}$.

3. Calculate the total distance travelled

Total distance $= |s(6) - s(0)| + |s(7) - s(6)|$
$= |-72 - 0| + \left|-67\frac{2}{3} - (-72)\right|$
$= 72 + 4\frac{1}{3}$
$= 76\frac{1}{3}\text{ m}$