题目

Problem

A particle moves in a straight line starting from a point $O$ before coming to instantaneous rest at a point $X$ . At time $t\,\mathrm{s}$ after leaving $O$ , the velocity $v\,\mathrm{m\,s^{-1}}$ of the particle is given by

\begin{align*} v =&\,7.2t^2 && 0\leqslant t\leqslant2,\\ v =&\,30.6-0.9t && 2\leqslant t\leqslant8,\\ v =&\,\frac{1600}{t^2}+kt && 8\leqslant t, \end{align*}

where $k$ is a constant. It is given that there is no instantaneous change in velocity at $t=8$ .

Find the distance $OX$ .

(9)

题目中文翻译

一个质点从点 $O$ 出发沿直线运动，之后在点 $X$ 瞬时静止。离开 $O$ 后 $t\,\mathrm{s}$ 时，质点的速度 $v\,\mathrm{m\,s^{-1}}$ 由下式给出：

\begin{align*} v =&\,7.2t^2 && 0\leqslant t\leqslant2,\\ v =&\,30.6-0.9t && 2\leqslant t\leqslant8,\\ v =&\,\frac{1600}{t^2}+kt && 8\leqslant t, \end{align*}

其中 $k$ 为常数。已知在 $t=8$ 时速度没有瞬时变化。

求距离 $OX$ 。

(9)

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

解法一

思路

展开

先由t等于8的速度连续求k，再求第三段第一次静止时刻。三段分别积分；检查此前速度非负，所以三段位移可以直接相加。

答题过程

展开

Continuity at $t=8$ gives

30.6-0.9(8)=\frac{1600}{8^2}+8k.

Thus $23.4=25+8k$ , so $k=-0.2$ . In the third interval, rest occurs when

\frac{1600}{t^2}-0.2t=0.

t^3=8000,\qquad t=20.

The first two velocity expressions are positive after departure; the third is positive for $8\le t<20$ . Hence $OX$ equals the sum of the three displacements:

D_1=\int_0^2 7.2t^2\,\mathrm{d}t =\big[2.4t^3\big]_0^2=19.2,

\begin{align*} D_2=&\,\int_2^8(30.6-0.9t)\,\mathrm{d}t\\[4mm] =&\,\big[30.6t-0.45t^2\big]_2^8\\[4mm] =&\,216-59.4=156.6. \end{align*}

\begin{align*} D_3=&\,\int_8^{20}(1600t^{-2}-0.2t)\,\mathrm{d}t\\[4mm] =&\,\big[-1600t^{-1}-0.1t^2\big]_8^{20}\\[4mm] =&\,-120-(-206.4)=86.4. \end{align*}

Therefore

\boxed{OX=19.2+156.6+86.4=262.2\,\mathrm{m}}.

CIE 9709 2023 November Paper 41 Q7

题目

解答

解法一

思路

答题过程