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CIE 9709 2026 June Paper 43 Q4

A Level / CIE / M1

CIE 9709 2026 June Paper 43 Paper · Question 4

题目

Problem

A train is travelling along a straight track with constant acceleration a m s−2a\,\mathrm{m\,s^{-2}}. It passes through points AA, BB and CC on the track in that order. The distances ABAB and BCBC are 280 m280\,\mathrm{m} and 170 m170\,\mathrm{m} respectively. The speed of the train at BB is 32 m s−132\,\mathrm{m\,s^{-1}}. The time it takes the train to travel from AA to BB is 2t s2t\,\mathrm{s} and from BB to CC is t st\,\mathrm{s}.

Find the value of aa and the value of tt.

(5)
题目中文翻译

一列火车以 a m s−2a\,\mathrm{m\,s^{-2}} 的恒定加速度沿直轨道行驶。它依次经过轨道上的点 AA、BB 和 CC。距离 ABAB 和 BCBC 分别为 280 m280\,\mathrm{m} 和 170 m170\,\mathrm{m}。火车在 BB 处的速率为 32 m s−132\,\mathrm{m\,s^{-1}}。火车从 AA 行驶至 BB 所需的时间为 2t s2t\,\mathrm{s},从 BB 行驶至 CC 所需的时间为 t st\,\mathrm{s}。

求 aa 和 tt 的值。

(5)
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

解法一

思路

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以B的速度作共同已知量。BC用初速度公式,AB用末速度公式,避免额外引入A的速度。

答题过程

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For the motion from BB to CC, using the initial speed at BB,

170=32t+12at2.(1)170=32t+\frac12at^2.\tag{1}

For the motion from AA to BB, using the final speed at BB,

280=32(2t)−12a(2t)2=64t−2at2.(2)280=32(2t)-\frac12a(2t)^2 =64t-2at^2.\tag{2}

Multiply (1) by 4 and add (2):

680+280= 128t+64t,960= 192t,t= 5.\begin{align*} 680+280=&\,128t+64t,\\[4mm] 960=&\,192t,\\[4mm] t=&\,\boxed{5}. \end{align*}

Substitution into (1) gives

170=160+12a(25),a=0.8.170=160+\frac12a(25), \qquad \boxed{a=0.8}.