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M1 2020 Oct Q8

A Level / Edexcel / M1

IAL 2020 Oct Paper · Question 8

题目

Problem

8.

The acceleration-time graph shown in Figure 5 represents part of a journey made by a car along a straight horizontal road. The car accelerated from rest at time t=0t=0.

(a) Find the distance travelled by the car during the first 4 s4\,\mathrm{s} of its journey.

(2)

(b) Find the total distance travelled by the car during the first 26 s26\,\mathrm{s} of its journey.

(6)
题目中文翻译

图 5 中的加速度—时间图像表示一辆汽车沿水平直路行驶的部分过程。汽车在 t=0t=0 时从静止开始加速。

(a) 求汽车在行程最初 4 s4\,\mathrm{s} 内行驶的路程。

(2)

(b) 求汽车在行程最初 26 s26\,\mathrm{s} 内行驶的总路程。

(6)

解答

(a)

解法一

思路

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加速度图像下的面积给速度变化,速度图像下的面积才给路程。先求 t=4t=4 的速度,再用对应速度图像的三角形面积。

答题过程

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During 0≤t≤40\leq t\leq4, the acceleration is 3 m s−23\,\mathrm{m\,s^{-2}}. Starting from rest, the speed at t=4t=4 is

v4=0+3(4)=12 m s−1.v_4=0+3(4)=12\,\mathrm{m\,s^{-1}}.

The velocity-time graph over this interval is a straight line from (0,0)(0,0) to (4,12)(4,12). Its triangular area gives the distance:

s1=12(4)(12)=24 m.s_1=\frac12(4)(12)=\boxed{24\,\mathrm{m}}.

(b)

解法一

思路

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每段加速度恒定,对应速度图像都是直线段,因此用梯形面积。最后一段虽然减速,速度仍然为正,路程可直接相加。

答题过程

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From (a), the speed at t=4t=4 is 12 m s−112\,\mathrm{m\,s^{-1}}. Between t=4t=4 and t=24t=24, the acceleration is 1.5 m s−21.5\,\mathrm{m\,s^{-2}}, so

v24=12+1.5(20)=42 m s−1.v_{24}=12+1.5(20)=42\,\mathrm{m\,s^{-1}}.

This part of the velocity-time graph is a straight line, whose trapezium area is

s2=12(12+42)(20)=540 m.s_2=\frac12(12+42)(20)=540\,\mathrm{m}.

During the last 22 seconds, the acceleration is −4 m s−2-4\,\mathrm{m\,s^{-2}}, so

v26=42−4(2)=34 m s−1.v_{26}=42-4(2)=34\,\mathrm{m\,s^{-1}}.

The corresponding trapezium area gives

s3=12(42+34)(2)=76 m.s_3=\frac12(42+34)(2)=76\,\mathrm{m}.

The velocity stays positive throughout, so the total distance is

s1+s2+s3=24+540+76=640 m.s_1+s_2+s_3=24+540+76=\boxed{640\,\mathrm{m}}.