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CIE 9709 2025 March Paper 42 Q2

A Level / CIE / M1

CIE 9709 2025 March Paper 42 Paper · Question 2

题目

Problem

A cyclist is travelling along a straight horizontal road at a speed of 4 m s−14\,\mathrm{m\,s^{-1}} when she passes a point OO. She accelerates at a constant rate for a distance of 42 m42\,\mathrm{m}, reaching a speed of V m s−1V\,\mathrm{m\,s^{-1}}. She maintains the speed of V m s−1V\,\mathrm{m\,s^{-1}} for 50 m50\,\mathrm{m} and then decelerates at 2 m s−22\,\mathrm{m\,s^{-2}} before coming to rest. The distance travelled while decelerating is 16 m16\,\mathrm{m}.

(a) Find the value of VV.

(2)

(b) Find the total time for which she is in motion from the instant that she passes OO.

(3)
题目中文翻译

一名骑车者沿水平直路行驶,以 4 m s−14\,\mathrm{m\,s^{-1}} 的速率经过点 OO。她以恒定加速度行驶 42 m42\,\mathrm{m},达到 V m s−1V\,\mathrm{m\,s^{-1}} 的速率。她保持 V m s−1V\,\mathrm{m\,s^{-1}} 的速率行驶 50 m50\,\mathrm{m},随后以 2 m s−22\,\mathrm{m\,s^{-2}} 的减速度减速直至停下。减速过程中行驶的路程为 16 m16\,\mathrm{m}。

(a) 求 VV 的值。

(2)

(b) 求从她经过 OO 的瞬间起,运动的总时间。

(3)
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

从最后减速段求进入该段时的速率,取正根。

答题过程

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For the deceleration stage, u=Vu=V, v=0v=0, a=−2a=-2 and s=16s=16. Hence

0=V2+2(−2)(16),V2=64.0=V^2+2(-2)(16),\qquad V^2=64.

Since VV is a speed,

V=8.\boxed{V=8}.

(b)

解法一

思路

展开

三段运动分别计算用时;匀加速段用平均速率,匀速段用路程除速率,最后再相加。

答题过程

展开

Let the three times be t1t_1, t2t_2 and t3t_3. During acceleration,

42=4+82t1,t1=7.42=\frac{4+8}{2}t_1, \qquad t_1=7.

At constant speed,

t2=508=254.t_2=\frac{50}{8}=\frac{25}{4}.

During deceleration,

0=8−2t3,t3=4.0=8-2t_3,\qquad t_3=4.

The total time is

7+254+4=694 s=17.25 s.7+\frac{25}{4}+4 =\boxed{\frac{69}{4}\,\mathrm{s}} =17.25\,\mathrm{s}.