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CIE 9709 2022 November Paper 43 Q4

A Level / CIE / M1

CIE 9709 2022 Nov Paper 43 Paper · Question 4

题目

Problem

A particle PP travels in the positive direction along a straight line with constant acceleration. PP travels a distance of 52 m52\,\mathrm{m} during the 2nd second of its motion and a distance of 64 m64\,\mathrm{m} during the 4th second of its motion.

(a) Find the initial speed and the acceleration of PP.

(5)

(b) Find the distance travelled by PP during the first 1010 seconds of its motion.

(2)

题目中文翻译

质点 PP 以恒定加速度沿直线的正方向运动。PP 在运动的第 22 秒内经过的路程为 52 m52\,\mathrm{m},在第 44 秒内经过的路程为 64 m64\,\mathrm{m}。

(a) 求 PP 的初始速率及加速度。

(5)

(b) 求 PP 在运动的前 1010 秒内经过的路程。

(2)

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

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第二秒和第四秒都是累计位移之差,分别用s(2)减s(1)、s(4)减s(3),联立求初速与加速度。

答题过程

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Let s(t)=ut+12at2s(t)=ut+\tfrac12at^2. The distances during the second and fourth seconds give

52=s(2)−s(1)=u+32a,52=s(2)-s(1)=u+\frac32a, 64=s(4)−s(3)=u+72a.64=s(4)-s(3)=u+\frac72a.

Subtracting,

12=2a,a=6 m s−2.12=2a,\qquad\boxed{a=6\,\mathrm{m\,s^{-2}}}.

Then

u=52−32(6)=43 m s−1.\boxed{u=52-\frac32(6)=43\,\mathrm{m\,s^{-1}}}.

(b)

解法一

思路

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把初速和加速度代入十秒累计位移,速度为正,所以就是路程。

答题过程

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The distance during the first 10 s10\,\mathrm{s} is

s=43(10)+12(6)(102)=430+300=730 m.s=43(10)+\frac12(6)(10^2) =430+300=\boxed{730\,\mathrm{m}}.