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CIE 9709 2024 November Paper 41 Q8

A Level / CIE / M1

CIE 9709 2024 Nov Paper 41 Paper · Question 8

题目

Problem

A particle PP moves in a straight line, passing through a point OO with velocity 4.2 m s−14.2\,\mathrm{m\,s^{-1}}. At time t st\,\mathrm{s} after PP passes OO, the acceleration, a m s−2a\,\mathrm{m\,s^{-2}}, of PP is given by a=0.6t−2.7a=0.6t-2.7.

Find the distance PP travels between the times at which it is at instantaneous rest.

(7)
题目中文翻译

一个质点 PP 沿直线运动,以 4.2 m s−14.2\,\mathrm{m\,s^{-1}} 的速度经过点 OO。PP 经过 OO 后 t st\,\mathrm{s} 时,其加速度 a m s−2a\,\mathrm{m\,s^{-2}} 由 a=0.6t−2.7a=0.6t-2.7 给出。

求 PP 在两次瞬时静止的时刻之间所运动的路程。

(7)
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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把加速度积分求速度,用初速确定常数。求两个静止时刻并检查其间速度符号;该段向负方向运动,所以路程取位移的相反数。

答题过程

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Integrating the acceleration,

v=0.3t2−2.7t+C.v=0.3t^2-2.7t+C.

Since v=4.2v=4.2 at t=0t=0, C=4.2C=4.2. Thus

v=0.3(t2−9t+14)=0.3(t−2)(t−7).v=0.3(t^2-9t+14)=0.3(t-2)(t-7).

The particle is instantaneously at rest at t=2t=2 and t=7t=7. For 2<t<72<t<7, v<0v<0, so the required distance is

D= −∫27(0.3t2−2.7t+4.2) dt= −[0.1t3−1.35t2+4.2t]27= 3.8−(−2.45)= 6.25 m.\begin{align*} D=&\,-\int_2^7(0.3t^2-2.7t+4.2)\,\mathrm{d}t\\[4mm] =&\,-\big[0.1t^3-1.35t^2+4.2t\big]_2^7\\[4mm] =&\,3.8-(-2.45)\\[4mm] =&\,\boxed{6.25\,\mathrm{m}}. \end{align*}