题目

Problem

A particle moves in a straight line. The velocity $v\,\mathrm{m\,s^{-1}}$ of the particle $t\,\mathrm{s}$ after leaving a fixed point $O$ is given by $v=k(20+pt-6t^2)$ , where $k$ and $p$ are constants. The acceleration of the particle at $t=1$ is $42\,\mathrm{m\,s^{-2}}$ , and the displacement of the particle from $O$ at $t=1$ is $93\,\mathrm{m}$ .

(a) Show that $k=3$ and $p=26$ .

(6)

(b) Find the distance moved by the particle between the time at which its acceleration is zero and the time at which its velocity is zero.

(5)

题目中文翻译

一个质点沿直线运动。离开固定点 $O$ 后 $t\,\mathrm{s}$ 时，质点的速度 $v\,\mathrm{m\,s^{-1}}$ 由 $v=k(20+pt-6t^2)$ 给出，其中 $k$ 和 $p$ 为常数。质点在 $t=1$ 时的加速度为 $42\,\mathrm{m\,s^{-2}}$ ，且此时相对于 $O$ 的位移为 $93\,\mathrm{m}$ 。

(a) 证明 $k=3$ 且 $p=26$ 。

(6)

(b) 求质点从加速度为零的时刻至速度为零的时刻所运动的路程。

(5)

原卷通用作答要求

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 $10\,\mathrm{m\,s^{-2}}$ 。

解答

(a)

解法一

思路

展开

速度微分给加速度，积分给从O出发的位移。两个已知条件建立联立方程，必须实际消元推出给定结论。

答题过程

展开

Differentiating the velocity gives

a=k(p-12t).

At $t=1$ , $a=42$ , so

kp-12k=42.\tag{1}

Since the particle starts at $O$ , its displacement is

\begin{align*} s=&\,\int_0^t k(20+p\tau-6\tau^2)\,\mathrm{d}\tau\\[4mm] =&\,k\bigg(20t+\frac p2t^2-2t^3\bigg). \end{align*}

Using $s=93$ at $t=1$ ,

18k+\frac{kp}{2}=93.\tag{2}

Twice (2) minus (1) gives

36k+kp-(kp-12k)=186-42,

so $48k=144$ and $k=3$ . Substitution into (1) gives $3p-36=42$ , hence

\boxed{k=3,\qquad p=26}.

(b)

解法一

思路

展开

先找加速度为零和速度为零的时刻，剔除负时间。检查该区间内速度为正，才能把位移增量当作路程。

答题过程

展开

With $k=3$ and $p=26$ ,

v=60+78t-18t^2,\qquad a=78-36t.

Thus $a=0$ when $t=13/6$ . Also

v=-6(3t+2)(t-5)=0,

giving $t=5$ or $t=-2/3$ . Only $t=5$ is after projection.

For $13/6\le t<5$ , the velocity is positive. Hence the required distance equals the increase in displacement:

\begin{align*} D=&\,\int_{13/6}^{5}(60+78t-18t^2)\,\mathrm{d}t\\[4mm] =&\,\big[60t+39t^2-6t^3\big]_{13/6}^{5}\\[4mm] =&\,525-\frac{4537}{18}\\[4mm] =&\,\boxed{\frac{4913}{18}\,\mathrm{m}} \approx273\,\mathrm{m}. \end{align*}

CIE 9709 2025 March Paper 42 Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程