质点 $P$ 沿直线运动。从直线上的点 $O$ 出发后 $t\,\mathrm{s}$ 时，$P$ 的加速度 $a\,\mathrm{m\,s^{-2}}$ 满足 $a=kt-3$，其中 $k$ 为正常数。在 $t=0$ 时，$P$ 的速度为 $1\,\mathrm{m\,s^{-1}}$。

(a) 已知 $P$ 从不瞬时静止，证明 $k>4.5$。

(b) 改为已知 $k=2.5$，求 $P$ 在时间区间 $1\leqslant t\leqslant3$ 内运动的总路程。

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 $10\,\mathrm{m\,s^{-2}}$。

先积分求速度。速度是开口向上的二次式，其最小值在正时间处；不曾静止要求判别式严格小于零。

Integrating $a=kt-3$ and using $v(0)=1$ gives

Since $k>0$, this quadratic opens upwards and its vertex occurs at $t=3/k>0$. Also $v(0)=1>0$, so if it had any real roots they would be positive: their sum is $6/k>0$ and product $2/k>0$.

Thus never being at instantaneous rest for $t\ge0$ requires no real roots:

Hence $\boxed{k>4.5}$, as required.

把速度配方，可直接看出正时间处的最小速度必须严格大于零。

Integrating the acceleration and using the initial velocity gives

Since $k>0$, the minimum occurs at the admissible time $t=3/k$. The particle starts with positive velocity and never reaches rest, so the minimum must be positive:

Multiplying by $2k>0$ gives $2k>9$, hence $\boxed{k>4.5}$.

也可令加速度为零求速度的极小值，用加速度的变号确认这是最小值。

The velocity is $v=kt^2/2-3t+1$. Its derivative is $a=kt-3$, which changes from negative to positive at $t=3/k$. Therefore the minimum velocity for $t\ge0$ is

Since the particle starts with positive velocity and is never at rest, $v_{\min}>0$. Thus $2k>9$, giving $\boxed{k>4.5}$.

速度在区间内的t=2变号。路程需要把变号前的负位移取反，再加变号后的正位移。

When $k=2.5$,

The zero-velocity times are $t=2/5$ and $t=2$; only $t=2$ lies in $[1,3]$. The velocity is negative on $[1,2)$ and positive on $(2,3]$.

An antiderivative is

Thus

Thus