题目

Problem

A particle starts from a point $O$ and moves in a straight line. The velocity $v\,\mathrm{m\,s^{-1}}$ of the particle at time $t\,\mathrm{s}$ after leaving $O$ is given by

v=k(3t^2-2t^3),

where $k$ is a constant.

(a) Verify that the particle returns to $O$ when $t=2$ .

(4)

(b) It is given that the acceleration of the particle is $-13.5\,\mathrm{m\,s^{-2}}$ for the positive value of $t$ at which $v=0$ .

Find $k$ and hence find the total distance travelled in the first two seconds of motion.

(6)

题目中文翻译

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

v=k(3t^2-2t^3),

其中 $k$ 为常数。

(a) 验证质点在 $t=2$ 时返回 $O$ 点。

(4)

(b) 已知在使 $v=0$ 的正值 $t$ 对应的时刻，质点的加速度为 $-13.5\,\mathrm{m\,s^{-2}}$ 。

求 $k$ ，并进而求运动的前两秒内经过的总路程。

(6)

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

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从出发点O量位移，用0到2的速度定积分验证净位移为零，不把积分当路程。

答题过程

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The displacement from $O$ at $t=2$ is

\begin{align*} s(2)=&\,\int_0^2 k(3t^2-2t^3)\,\mathrm{d}t\\[4mm] =&\,k\bigg[t^3-\frac12t^4\bigg]_0^2\\[4mm] =&\,k(8-8)=0. \end{align*}

Thus the particle is back at $O$ when $\boxed{t=2}$ .

(b)

解法一

思路

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给定非零加速度使k不为零，速度的正零点为1.5。微分求k后，承接(a)位移原函数，先求转折点到O的距离，再加上返回路程。

答题过程

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Since the given acceleration is nonzero, $k\ne0$ . The positive zero of velocity satisfies

k t^2(3-2t)=0, \qquad t=\frac32.

Differentiating,

a=k(6t-6t^2).

At $t=3/2$ ,

-13.5=k(9-13.5)=-4.5k, \qquad\boxed{k=3}.

The velocity is positive before $3/2$ and negative afterwards until $2$ . From (a), the displacement at the turning point is

\begin{align*} s\bigg(\frac32\bigg)=&\,3\bigg[\bigg(\frac32\bigg)^3-\frac12\bigg(\frac32\bigg)^4\bigg]\\[4mm] =&\,3\bigg(\frac{27}{8}-\frac{81}{32}\bigg)\\[4mm] =&\,\frac{81}{32}\,\mathrm{m}. \end{align*}

Since the particle returns to $O$ at $t=2$ , its total distance is

\boxed{2\bigg(\frac{81}{32}\bigg)=\frac{81}{16}\,\mathrm{m}} \approx5.06\,\mathrm{m}.

CIE 9709 2022 June Paper 41 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程