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CIE 9709 2025 November Paper 43 Q1

A Level / CIE / M1

CIE 9709 2025 Nov Paper 43 Paper · Question 1

题目

Problem

The diagram shows the velocity-time graph for the motion of an athlete. The athlete runs in a straight line from point AA to point BB, runs back to point AA and finishes at point BB. The graph consists of four straight line segments.

(a) Find the deceleration of the athlete 77 seconds after leaving AA.

(1)

(b) Find the total distance travelled by the athlete.

(2)
题目中文翻译

图中显示一名运动员运动的速度—时间图像。运动员从点 AA 沿直线跑到点 BB,再跑回点 AA,最后在点 BB 结束。图像由四条直线段组成。

(a) 求运动员离开 AA 后 77 秒时的减速度。

(1)

(b) 求运动员运动的总路程。

(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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7秒位于5至15秒的下降直线段,斜率是加速度,减速度取大小。

答题过程

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At t=7t=7, the acceleration is the gradient of the segment from (5,8)(5,8) to (15,−8)(15,-8):

a=−8−815−5=−1.6 m s−2.a=\frac{-8-8}{15-5}=-1.6\text{ m s}^{-2}.

Thus the deceleration is 1.6 m s−2\boxed{1.6\text{ m s}^{-2}}.

(b)

解法一

思路

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每段往返的路程是对应三角形面积大小,负速度面积也按正值相加。

答题过程

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The graph forms three triangles, each with base 10 seconds and height 8 m s−18\,\mathrm{m\,s^{-1}} in magnitude. The total distance is

3(12(10)(8))=120 m.3\biggl(\frac12(10)(8)\biggr)=\boxed{120\text{ m}}.