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CIE 9709 2022 June Paper 41 Q1

A Level / CIE / M1

CIE 9709 2022 June Paper 41 Paper · Question 1

题目

Problem

A car starts from rest and moves in a straight line with constant acceleration for a distance of 200 m200\,\mathrm{m}, reaching a speed of 25 m s−125\,\mathrm{m\,s^{-1}}. The car then travels at this speed for 400 m400\,\mathrm{m}, before decelerating uniformly to rest over a period of 5 s5\,\mathrm{s}.

(a) Find the time for which the car is accelerating.

(2)

(b) Sketch the velocity–time graph for the motion of the car, showing the key points.

(2)

(c) Find the average speed of the car during its motion.

(2)

题目中文翻译

一辆汽车从静止出发,以恒定加速度沿直线行驶 200 m200\,\mathrm{m},达到 25 m s−125\,\mathrm{m\,s^{-1}} 的速率。随后汽车以此速率行驶 400 m400\,\mathrm{m},再经过 5 s5\,\mathrm{s} 匀减速至静止。

(a) 求汽车加速所用的时间。

(2)

(b) 画出汽车运动的速度—时间图的草图,并标出关键点。

(2)

(c) 求汽车整个运动过程中的平均速率。

(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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匀加速段平均速率为初末速率平均值,路程除平均速率即可求时间。

答题过程

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For constant acceleration,

200=0+252t,t=16 s.200=\frac{0+25}{2}t, \qquad\boxed{t=16\,\mathrm{s}}.

(b)

解法一

思路

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匀速400米用时16秒,所以平台终点在32秒,再经5秒减速到37秒。

答题过程

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The constant-speed stage lasts 400/25=16 s400/25=16\,\mathrm{s}. Therefore sketch straight segments joining

(0,0),(16,25),(32,25),(37,0)(0,0),\quad(16,25),\quad(32,25),\quad(37,0)

on a velocity-time graph, with tt in seconds and vv in m s−1\mathrm{m\,s^{-1}}.

(c)

解法一

思路

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平均速率是总路程除总时间,减速段路程用三角形面积。

答题过程

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The total distance is

D=200+400+12(25)(5)=662.5 m.D=200+400+\frac12(25)(5)=662.5\,\mathrm{m}.

The total time is 37 s37\,\mathrm{s}, so

average speed=662.537≈17.9 m s−1(3 s.f.).\boxed{\text{average speed}=\frac{662.5}{37} \approx17.9\,\mathrm{m\,s^{-1}}}\quad\text{(3 s.f.)}.