质点 $P$ 从点 $O$ 出发，沿直线运动。离开 $O$ 后 $t\,\mathrm{s}$ 时，$P$ 的速度 $v\,\mathrm{m\,s^{-1}}$ 由 $v=\frac13(2t-3)(t-4)$ 给出。

(a) 求 $P$ 在 $t=2$ 时的加速度。

(b) 求 $P$ 在运动的前 $3$ 秒内运动的总路程。

(c) 判断 $P$ 是否返回 $O$。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$。

展开速度后求导，再代入2。

At $t=2$,

前3秒内只有1.5秒反向，把两个区间位移的大小相加。

The velocity is zero at $t=1.5$ and $t=4$. It is positive for $0\leq t<1.5$ and negative for $1.5<t\leq3$.

Since $P$ starts at $O$, its displacement is

In particular,

Thus the total distance is

在4秒后速度恢复为正；返回O的最后机会在4秒之前，检查转向点位移仍为正。

After first moving away from $O$, $P$ moves back towards it only during $1.5<t<4$. At the end of that interval,

For $t>4$, $v>0$ and the displacement increases again. Thus $P$ does not return to $O$.

也可令位移为零，除去出发时刻t=0后用判别式排除其余实根。

For a return to $O$ at $t>0$,

Hence $4t^2-33t+72=0$ would be required. Its discriminant is

There is no real solution, so $P$ never returns to $O$ after departure.