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CIE 9709 2026 March Paper 42 Q7

A Level / CIE / M1

CIE 9709 2026 March Paper 42 Paper · Question 7

题目

Problem

A particle PP travels in a straight line. The velocity of PP at time t st\,\mathrm{s} is v m s−1v\,\mathrm{m\,s^{-1}}, where

v=t−7t12+12.v=t-7t^{\frac12}+12.

(a) Find the acceleration of PP at t=4t=4.

(2)

(b) Find the values of tt at which PP is instantaneously at rest.

(3)

(c) Find the total distance travelled by PP in the interval 0⩽t⩽250\leqslant t\leqslant25.

(5)
题目中文翻译

质点 PP 沿直线运动。在时刻 t st\,\mathrm{s},PP 的速度为 v m s−1v\,\mathrm{m\,s^{-1}},其中

v=t−7t12+12.v=t-7t^{\frac12}+12.

(a) 求 PP 在 t=4t=4 时的加速度。

(2)

(b) 求 PP 瞬时静止时 tt 的值。

(3)

(c) 求 PP 在区间 0⩽t⩽250\leqslant t\leqslant25 内运动的总路程。

(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

(a)

解法一

思路

展开

根号项求导时幂次减一,代入4保留加速度符号。

答题过程

展开 a=dvdt=1−72t−1/2.a=\frac{\mathrm{d}v}{\mathrm{d}t}=1-\frac72t^{-1/2}.

At t=4t=4,

a=1−724=1−74=−0.75 m s−2.a=1-\frac7{2\sqrt4}=1-\frac74 =\boxed{-0.75\text{ m s}^{-2}}.

(b)

解法一

思路

展开

以√t为变量把速度因式分解,根号为3或4,时间需平方。

答题过程

展开

At instantaneous rest,

t−7t+12=(t−3)(t−4)=0.t-7\sqrt t+12=(\sqrt t-3)(\sqrt t-4)=0.

Thus t=3\sqrt t=3 or 4, giving

t=9 sort=16 s.\boxed{t=9\text{ s}\quad\text{or}\quad t=16\text{ s}}.

(c)

解法一

思路

展开

9、16秒处反向,分别计算三个区间位移,再把绝对值相加。

答题过程

展开

The factorisation in (b) shows that the velocity is positive on [0,9)[0,9), negative on (9,16)(9,16) and positive on (16,25](16,25].

An antiderivative of the velocity is

S(t)=12t2−143t3/2+12t.S(t)=\frac12t^2-\frac{14}{3}t^{3/2}+12t.

Its values are

S(0)= 0,S(9)= 452,S(16)= 643,S(25)= 1756.\begin{align*} S(0)=&\,0,\\[4mm] S(9)=&\,\frac{45}{2},\\[4mm] S(16)=&\,\frac{64}{3},\\[4mm] S(25)=&\,\frac{175}{6}. \end{align*}

Therefore the total distance is

d= [S(9)−S(0)]+[S(9)−S(16)] +[S(25)−S(16)]= 452+76+476=31.5 m.\begin{align*} d=&\,\bigl[S(9)-S(0)\bigr]+\bigl[S(9)-S(16)\bigr]\\[4mm] &\,\hspace{2pt}+\bigl[S(25)-S(16)\bigr]\\[4mm] =&\,\frac{45}{2}+\frac76+\frac{47}{6} =\boxed{31.5\text{ m}}. \end{align*}