一个质点 $X$ 沿直线运动，在离开定点 $O$ 之后 $t$ 秒时，其速度 $v$ m/s 由 $v = \frac{1}{2}t^2 - t - 4$ 给出。

1. Find the time when the velocity is minimum.

To find the minimum velocity, we can differentiate $v$ with respect to $t$ and equate to zero:
$v = \frac{1}{2}t^2 - t - 4$
$\frac{dv}{dt} = t - 1$
For minimum velocity, $\frac{dv}{dt} = 0$
$t - 1 = 0 \implies t = 1$

2. Find the general expression for displacement $s$ by integrating velocity.

$s = \int v \, dt$
$s = \int \left(\frac{1}{2}t^2 - t - 4\right) \, dt$
$s = \frac{1}{6}t^3 - \frac{1}{2}t^2 - 4t + c$

3. Determine the constant of integration using the initial condition.

Since the particle leaves a fixed point $O$, its displacement is $0$ when $t = 0$.
$0 = \frac{1}{6}(0)^3 - \frac{1}{2}(0)^2 - 4(0) + c \implies c = 0$
So, $s = \frac{1}{6}t^3 - \frac{1}{2}t^2 - 4t$

4. Calculate the displacement at the time of minimum velocity.

Substitute $t = 1$ into the displacement equation:
$s = \frac{1}{6}(1)^3 - \frac{1}{2}(1)^2 - 4(1)$
$s = \frac{1}{6} - \frac{1}{2} - 4$
$s = -\frac{1}{3} - 4 = -4\frac{1}{3}$

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

$v = \frac{1}{2}t^2 - t - 4 = 0$

2. Solve the quadratic equation for $t$ and reject the negative root

$t^2 - 2t - 8 = 0$
$(t - 4)(t + 2) = 0$
$t = 4 \text{ or } t = -2$
Since $t \ge 0$, $t = 4$.

1. Find the time when the particle turns (from part b)

The particle turns when $v = 0$.
$\frac{1}{2}t^2 - t - 4 = 0$
$t^2 - 2t - 8 = 0$
$(t - 4)(t + 2) = 0$
Since $t \ge 0$, $t = 4$.

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

$s = \int (\frac{1}{2}t^2 - t - 4) \, dt$
$s = \frac{1}{6}t^3 - \frac{1}{2}t^2 - 4t + C$
Since the particle leaves from $O$, $s = 0$ when $t = 0$, so $C = 0$.
$s(t) = \frac{1}{6}t^3 - \frac{1}{2}t^2 - 4t$

3. Calculate the displacement at $t=0$, $t=4$, and $t=6$

$s(0) = 0$
$s(4) = \frac{1}{6}(4)^3 - \frac{1}{2}(4)^2 - 4(4) = \frac{32}{3} - 8 - 16 = -\frac{40}{3}$
$s(6) = \frac{1}{6}(6)^3 - \frac{1}{2}(6)^2 - 4(6) = 36 - 18 - 24 = -6$

4. Calculate the total distance travelled

Distance from $t=0$ to $t=4$: $|-\frac{40}{3} - 0| = \frac{40}{3}$
Distance from $t=4$ to $t=6$: $|-6 - (-\frac{40}{3})| = |-6 + 13\frac{1}{3}| = \frac{22}{3}$
Total distance = $\frac{40}{3} + \frac{22}{3} = \frac{62}{3} = 20\frac{2}{3}$ m