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CIE 9709 2026 June Paper 42 Q1

A Level / CIE / M1

CIE 9709 2026 June Paper 42 Paper · Question 1

题目

Problem

A car of mass 960 kg960\,\mathrm{kg} is moving on a straight horizontal road. There is a constant force of magnitude 620 N620\,\mathrm{N} resisting the motion of the car.

(a) Calculate the power developed by the engine of the car when it is moving at a constant speed of 25 m s−125\,\mathrm{m\,s^{-1}}.

(2)

(b) Given that the power is suddenly increased by 12 kW12\,\mathrm{kW}, find the instantaneous acceleration of the car.

(2)
题目中文翻译

一辆质量为 960 kg960\,\mathrm{kg} 的汽车在一条水平直路上行驶。汽车受到大小恒为 620 N620\,\mathrm{N}、阻碍其运动的力。

(a) 计算汽车以 25 m s−125\,\mathrm{m\,s^{-1}} 的恒定速率行驶时发动机输出的功率。

(2)

(b) 已知功率突然增加了 12 kW12\,\mathrm{kW},求汽车的瞬时加速度。

(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

展开

匀速水平运动,牵引力恰好抵消阻力,再用P=Fv。

答题过程

展开

At constant speed the resultant force is zero, so the driving force is 620 N620\,\mathrm{N}. Thus

P=Fv=620(25)=15500 W=15.5 kW.P=Fv=620(25)=15500\text{ W}=\boxed{15.5\text{ kW}}.

(b)

解法一

思路

展开

功率增加12kW而非变为12kW;瞬间速度仍25,求新牵引力后用F=ma。

答题过程

展开

The new power is 15500+12000=27500 W15500+12000=27500\,\mathrm{W}. The speed is still 25 m s−125\,\mathrm{m\,s^{-1}} at that instant, so

F=2750025=1100 N.F=\frac{27500}{25}=1100\text{ N}.

Therefore

1100−620=960a,a=0.5 m s−2.1100-620=960a, \qquad a=\boxed{0.5\text{ m s}^{-2}}.