题目

Problem

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

v=t^3-\frac{9}{2}t^2+1\text{ for }0\leqslant t\leqslant4.

You may assume that the velocity of the particle is positive for $t<\frac12$ , is zero at $t=\frac12$ and is negative for $t>\frac12$ .

(a) Find the distance travelled between $t=0$ and $t=\frac12$ .

(4)

(b) Find the positive value of $t$ at which the acceleration is zero. Hence find the total distance travelled between $t=0$ and this instant.

(4)

题目中文翻译

一个质点从点 $O$ 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=t^3-\frac{9}{2}t^2+1\text{，其中 }0\leqslant t\leqslant4.

可以假设质点的速度在 $t<\frac12$ 时为正，在 $t=\frac12$ 时为零，在 $t>\frac12$ 时为负。

(a) 求从 $t=0$ 到 $t=\frac12$ 经过的路程。

(4)

(b) 求加速度为零时 $t$ 的正值。由此求从 $t=0$ 到该时刻经过的总路程。

(4)

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

给定前半秒速度为正，路程直接用定积分，不需取负号。

答题过程

展开

Since the velocity is positive on the stated interval,

\begin{align*} D=&\,\int_0^{1/2}\bigg(t^3-\frac92t^2+1\bigg)\,\mathrm{d}t\\[4mm] =&\,\bigg[\frac14t^4-\frac32t^3+t\bigg]_0^{1/2}\\[4mm] =&\,\frac1{64}-\frac3{16}+\frac12\\[4mm] =&\,\boxed{\frac{21}{64}\,\mathrm{m}}. \end{align*}

(b)

解法一

思路

展开

先求加速度零点，取正时间3。承接(a)的路程与位移原函数，后段速度负，所以后段路程取位移的相反数。

答题过程

展开

Differentiating,

a=3t^2-9t=3t(t-3).

Thus $a=0$ at $t=0$ and $t=3$ . The positive time is $\boxed{3\,\mathrm{s}}$ .

Using the antiderivative from (a), define

S(t)=\frac14t^4-\frac32t^3+t.

Then $S(1/2)=21/64$ and

S(3)=\frac{81}{4}-\frac{81}{2}+3=-\frac{69}{4}.

Since $v<0$ from $t=1/2$ to $t=3$ , the total distance is

\begin{align*} D=&\,\frac{21}{64}-\bigg[S(3)-S\bigg(\frac12\bigg)\bigg]\\[4mm] =&\,\frac{21}{64}+\frac{69}{4}+\frac{21}{64}\\[4mm] =&\,\boxed{\frac{573}{32}\,\mathrm{m}} \approx17.9\,\mathrm{m}. \end{align*}

CIE 9709 2024 March Paper 42 Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程