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CIE 9709 2025 June Paper 45 Q3

A Level / CIE / M1

CIE 9709 2025 June Paper 45 Paper · Question 3

题目

Problem

A train travels 4.8 km4.8\,\mathrm{km} between two stations, AA and BB. The train starts from rest at AA and accelerates at a constant rate until it reaches a speed of 30 m s−130\,\mathrm{m\,s^{-1}}. It then travels at this constant speed for TT seconds, before decelerating at a constant rate, coming to rest at BB. The total time for the journey is 180 s180\,\mathrm{s}.

Find the value of TT and hence find the distance moved by the train while travelling at the constant speed of 30 m s−130\,\mathrm{m\,s^{-1}}.

(5)

题目中文翻译

一列火车在两个车站 AA 和 BB 之间行驶 4.8 km4.8\,\mathrm{km}。火车从 AA 站由静止开始以恒定加速度加速,直到速率达到 30 m s−130\,\mathrm{m\,s^{-1}}。随后火车以此恒定速率行驶 TT 秒,然后以恒定减速度减速,在 BB 站停下。全程所需时间为 180 s180\,\mathrm{s}。

求 TT 的值,并由此求火车以 30 m s−130\,\mathrm{m\,s^{-1}} 的恒定速率行驶时所经过的距离。

(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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两段匀变速运动的平均速率都是15米每秒,但两段时间不必相等。利用它们的时间总和180−T列总路程。

答题过程

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Let the acceleration and deceleration times be t1,t3t_1,t_3. Since each of these stages has average speed (0+30)/2=15 m s−1(0+30)/2=15\,\mathrm{m\,s^{-1}},

s1= 15t1,s3= 15t3,t1+t3= 180−T.\begin{align*} s_1=&\,15t_1,\\[4mm] s_3=&\,15t_3,\\[4mm] t_1+t_3=&\,180-T. \end{align*}

Adding the three distances and converting 4.8 km4.8\,\mathrm{km} to metres,

4800= 15(t1+t3)+30T= 15(180−T)+30T,4800= 2700+15T.\begin{align*} 4800=&\,15(t_1+t_3)+30T \\[4mm] =&\,15(180-T)+30T, \\[4mm] 4800=&\,2700+15T. \end{align*}

Thus T=140\boxed{T=140}. Hence the distance travelled at constant speed is

30(140)=4200 m.30(140)=\boxed{4200\,\mathrm{m}}.

解法二

思路

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速度图是梯形,上底为T,下底为180,高为30;梯形面积就是总路程。

答题过程

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The velocity-time graph is a trapezium with parallel sides of lengths TT and 180180, and height 3030. Its area is the total distance, so

12(T+180)(30)=4800.\frac12(T+180)(30)=4800.

Thus T+180=320T+180=320, giving T=140\boxed{T=140}. Hence the distance at constant speed is

30T=4200 m.30T=\boxed{4200\,\mathrm{m}}.