题目

Problem

A particle $P$ moves in a straight line. The velocity $v\,\mathrm{m\,s^{-1}}$ at time $t$ seconds is given by

\begin{align*} v=&\,0.5t &&\text{for }0\le t\le10,\\ v=&\,0.25t^2-8t+60 &&\text{for }10\le t\le20. \end{align*}

(a) Show that there is an instantaneous change in the acceleration of the particle at $t=10$ .

(3)

(b) Find the total distance covered by $P$ in the interval $0\le t\le20$ .

(6)

题目中文翻译

质点 $P$ 沿直线运动。在 $t$ 秒时的速度 $v\,\mathrm{m\,s^{-1}}$ 由下式给出：

\begin{align*} v=&\,0.5t &&\text{当 }0\le t\le10,\\ v=&\,0.25t^2-8t+60 &&\text{当 }10\le t\le20. \end{align*}

(a) 证明质点的加速度在 $t=10$ 时发生瞬时变化。

(3)

(b) 求 $P$ 在区间 $0\le t\le20$ 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

展开

对两段速度分别求导，比较t=10左、右两侧的加速度。

答题过程

展开

For $0<t<10$ ,

a=\frac{\mathrm{d}v}{\mathrm{d}t}=0.5\,\mathrm{m\,s^{-2}}.

For $10<t<20$ ,

a=\frac{\mathrm{d}v}{\mathrm{d}t}=0.5t-8.

The acceleration immediately after $t=10$ is $0.5(10)-8=-3\,\mathrm{m\,s^{-2}}$ , whereas immediately before it is $0.5\,\mathrm{m\,s^{-2}}$ . These unequal one-sided values show an instantaneous change in acceleration at $t=10$ .

(b)

解法一

思路

展开

先找速度变号时刻，再分段累加路程；负速度区间的位移要取相反数。

答题过程

展开

In the first $10$ seconds the distance is

\int_0^{10}0.5t\,\mathrm{d}t =\big[0.25t^2\big]_0^{10}=25\,\mathrm{m}.

For the second interval,

\begin{align*} v=&\,\frac14(t^2-32t+240)\\[4mm] =&\,\frac14(t-12)(t-20). \end{align*}

Thus $v\ge0$ on $10\le t\le12$ and $v\le0$ on $12\le t\le20$ . An antiderivative is

F(t)=\frac{t^3}{12}-4t^2+60t.

The two distances in this interval are

\begin{align*} F(12)-F(10)=&\,288-\frac{850}{3}\\[4mm] =&\,\frac{14}{3},\\[4mm] F(12)-F(20)=&\,288-\frac{800}{3}\\[4mm] =&\,\frac{64}{3}. \end{align*}

Hence the total distance is

25+\frac{14}{3}+\frac{64}{3}=\boxed{51\,\mathrm{m}}.

CIE 9709 2022 June 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