一质点在直线上运动，使得在通过固定点 O 后 $t$ 秒，其速度 $v$ m/s 由 $v = t^2 + kt + 5$ 给出，其中 k 为常数。该质点先在 A 处瞬时静止，然后在 B 处瞬时静止。

1. Find the acceleration from the velocity expression and use that the particle is at O when $t=0$.

Given $v=t^2+kt+5$, the acceleration is
$a=\frac{dv}{dt}=2t+k.$
At point O, $t=0$, and $a=-6$.
$2(0)+k=-6$
$k=-6$

1. Use $k=-6$ and find the times when the particle is at rest.

From part (i), $k=-6$.

So
$v=t^2-6t+5.$

At instantaneous rest, $v=0$:
$t^2-6t+5=0$
$(t-1)(t-5)=0.$

Hence
$t=1 \quad \text{or} \quad t=5.$

2. Set up the displacement between A and B as the integral of velocity from $t=1$ to $t=5$.

The displacement from A to B is
$\int_1^5 v\,dt=\int_1^5 (t^2-6t+5)\,dt.$

3. Integrate the velocity expression and evaluate the displacement.

$\int (t^2-6t+5)\,dt=\frac{t^3}{3}-3t^2+5t.$

So
$\int_1^5 (t^2-6t+5)\,dt$
$=\left[\frac{t^3}{3}-3t^2+5t\right]_1^5$
$=\left(\frac{125}{3}-75+25\right)-\left(\frac{1}{3}-3+5\right)$
$=-\frac{25}{3}-\frac{7}{3}$
$=-\frac{32}{3}.$

4. Convert displacement to distance by taking the magnitude.

The displacement from A to B is negative because the particle moves in the negative direction between $t=1$ and $t=5$.

Therefore the distance between A and B is
$\left|-\frac{32}{3}\right|=\frac{32}{3}\text{ m}.$

1. Identify the time intervals over which the direction of motion is constant.

From earlier parts, $k=-6$, so
$v=t^2-6t+5=(t-1)(t-5).$
The particle is instantaneously at rest when $v=0$, so
$t=1 \quad \text{or} \quad t=5.$
Hence split the first 6 seconds into the intervals
$0\leq t\leq 1,\quad 1\leq t\leq 5,\quad 5\leq t\leq 6.$

2. Find the displacements over the required intervals.

The displacement from time $a$ to time $b$ is
$\int_a^b v\,dt=\int_a^b (t^2-6t+5)\,dt.$
An antiderivative is
$\int (t^2-6t+5)\,dt=\frac{t^3}{3}-3t^2+5t.$
Let
$s(t)=\frac{t^3}{3}-3t^2+5t.$
Then
$s(0)=0,$
$s(1)=\frac{1}{3}-3+5=\frac{7}{3},$
$s(5)=\frac{125}{3}-75+25=-\frac{25}{3},$
$s(6)=72-108+30=-6.$

3. Add the magnitudes of the displacements to obtain total distance.

Total distance travelled from $t=0$ to $t=6$ is
$|s(1)-s(0)|+|s(5)-s(1)|+|s(6)-s(5)|.$
So
$\text{Total distance}=\left|\frac{7}{3}-0\right|+\left|-\frac{25}{3}-\frac{7}{3}\right|+\left|-6-\left(-\frac{25}{3}\right)\right|.$
$=\frac{7}{3}+\frac{32}{3}+\frac{7}{3}$
$=\frac{46}{3}\text{ m}.$

1. Find the time when the velocity is minimum.

Using $k=-6$ from part (i),
$v=t^2-6t+5.$
Complete the square:
$v=(t-3)^2-4.$
Hence the minimum velocity occurs when
$t=3.$

2. Find the position of C and compare its distance from O and B.

The displacement from $O$ to $C$ is
$s_C=\int_0^3 (t^2-6t+5)\,dt$
$=\left[\frac{t^3}{3}-3t^2+5t\right]_0^3$
$=9-27+15=-3.$
So $OC=3$ m.

From earlier parts, $A$ is at $s=\frac{7}{3}$ and $AB=\frac{32}{3}$, so
$s_B=\frac{7}{3}-\frac{32}{3}=-\frac{25}{3}.$
Thus
$BC=\left|-\frac{25}{3}-(-3)\right|=\left|-\frac{16}{3}\right|=\frac{16}{3}\text{ m}.$
Since
$3<\frac{16}{3},$
$C$ is nearer to $O$.