Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2023 November Paper 43 Q6

A Level / CIE / M1

CIE 9709 2023 Nov Paper 43 Paper · Question 6

题目

Problem

A particle moves in a straight line. At time t st\,\mathrm{s}, the acceleration, a m s−2a\,\mathrm{m\,s^{-2}}, of the particle is given by a=36−6ta=36-6t. The velocity of the particle is 27 m s−127\,\mathrm{m\,s^{-1}} when t=2t=2.

(a) Find the values of tt when the particle is at instantaneous rest.

(4)

(b) Find the total distance the particle travels during the first 1212 seconds.

(4)

题目中文翻译

一个质点沿直线运动。在时刻 t st\,\mathrm{s},质点的加速度 a m s−2a\,\mathrm{m\,s^{-2}} 由 a=36−6ta=36-6t 给出。当 t=2t=2 时,质点的速度为 27 m s−127\,\mathrm{m\,s^{-1}}。

(a) 求质点瞬时静止时 tt 的各个值。

(4)

(b) 求质点在前 1212 秒内经过的总路程。

(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 (gg) is needed, use 10 m s−210\,\mathrm{m\,s^{-2}}.

需要使用重力加速度 gg 的数值时,取 10 m s−210\,\mathrm{m\,s^{-2}}。

解答

(a)

解法一

思路

展开

加速度积分求速度,用t等于2时速度27确定常数,再解两个静止时刻。

答题过程

展开

Integrating,

v=36t−3t2+C.v=36t-3t^2+C.

Using v=27v=27 at t=2t=2,

27=72−12+C,C=−33.27=72-12+C,\qquad C=-33.

Thus

v=−3t2+36t−33=−3(t−1)(t−11).v=-3t^2+36t-33=-3(t-1)(t-11).

Therefore the particle is instantaneously at rest at

t=1 and t=11.\boxed{t=1\text{ and }t=11}.

(b)

解法一

思路

展开

速度在两个静止时刻变号,将0到12分为三段,各段位移取绝对值后相加。

答题过程

展开

From (a), the velocity is negative on 0≤t<10\le t<1, positive on 1<t<111<t<11, and negative on 11<t≤1211<t\le12. An antiderivative is

S(t)=18t2−t3−33t.S(t)=18t^2-t^3-33t.

Its values are

S(0)= 0,S(1)= −16,S(11)= 484,S(12)= 468.\begin{align*} S(0)=&\,0,\\[4mm] S(1)=&\,-16,\\[4mm] S(11)=&\,484,\\[4mm] S(12)=&\,468. \end{align*}

Hence

D= −[S(1)−S(0)]+[S(11)−S(1)] −[S(12)−S(11)]= 16+500+16= 532 m.\begin{align*} D=&\,-[S(1)-S(0)]+[S(11)-S(1)]\\[4mm] &\,\hspace{2pt}-[S(12)-S(11)]\\[4mm] =&\,16+500+16\\[4mm] =&\,\boxed{532\,\mathrm{m}}. \end{align*}