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M1 2026 Jan A Q5

A Level / Edexcel / M1

IAL 2026 Jan A Paper · Question 5

题目

Problem

5.

Two posts, AA and BB, are fixed at the side of a straight horizontal road and are 816 m816\,\mathrm{m} apart, as shown in Figure 1. A car and a van are at rest side by side on the road and level with AA. At time t=0t=0, the car and the van start to move in the direction ABAB.

The car accelerates from rest with constant acceleration for TT seconds until it reaches a speed of 24 m s−124\,\mathrm{m\,s^{-1}} The car then moves at a constant speed of 24 m s−124\,\mathrm{m\,s^{-1}}

The van accelerates from rest with constant acceleration for 12 s12\,\mathrm{s} until it reaches a speed of V m s−1V\,\mathrm{m\,s^{-1}} The van then moves at a constant speed of V m s−1V\,\mathrm{m\,s^{-1}}

When the car has been moving at 24 m s−124\,\mathrm{m\,s^{-1}} for 30 s30\,\mathrm{s}, the van draws level with the car at BB, and each vehicle has then travelled a distance of 816 m816\,\mathrm{m}.

(a) Sketch, on the same diagram, a speed-time graph for the motion of each vehicle from AA to BB.

(3)

(b) Find the value of TT.

(3)

(c) Find the value of VV.

(3)
题目中文翻译

如图 1 所示,两根柱子 AA 和 BB 固定在一条水平直路旁,相距 816 m816\,\mathrm{m}。 一辆汽车与一辆货车并排静止在公路上与 AA 齐平的位置。 在 t=0t=0 时,两车开始沿 ABAB 方向运动。

汽车由静止以恒定加速度行驶 TT 秒,直到速率达到 24 m s−124\,\mathrm{m\,s^{-1}}。 随后汽车以恒定速率 24 m s−124\,\mathrm{m\,s^{-1}} 行驶。

货车由静止以恒定加速度行驶 12 s12\,\mathrm{s},直到速率达到 V m s−1V\,\mathrm{m\,s^{-1}}。 随后货车以恒定速率 V m s−1V\,\mathrm{m\,s^{-1}} 行驶。

当汽车已以 24 m s−124\,\mathrm{m\,s^{-1}} 行驶 30 s30\,\mathrm{s} 时,货车在 BB 与汽车齐平,此时两车各已行驶 816 m816\,\mathrm{m}。

(a) 在同一图中画出两车从 AA 行驶至 BB 的速率—时间图像草图。

(3)

(b) 求 TT 的值。

(3)

(c) 求 VV 的值。

(3)

解答

(a)

解法一

思路

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两车均从静止开始,先画斜线加速段,再画水平匀速段。两图到达 BB 的时间相同,但此时两车并未停车,不能把曲线画回时间轴。

答题过程

展开

Use the same axes for both graphs, labelled time t (s)t\text{ (s)} and speed v (m s−1)v\text{ (m s}^{-1}\text{)}.

For the car, draw a straight line from (0,0)(0,0) to (T,24)(T,24), followed by a horizontal line to (T+30,24)(T+30,24).

For the van, draw a straight line from (0,0)(0,0) to (12,V)(12,V), followed by a horizontal line to (T+30,V)(T+30,V).

The van’s graph crosses the car’s graph during the journey. Draw T<12T<12 and V>24V>24, and label the car, van, TT, 1212, T+30T+30, 2424 and VV. Both graphs end at t=T+30t=T+30 without dropping to zero.

(b)

解法一

思路

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汽车的总时间是 T+30T+30,而非 3030 秒。将汽车的三角形加速面积与长方形匀速面积相加。

答题过程

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The distance travelled by the car is the area under its speed–time graph:

12T(24)+30(24)=816.\frac12T(24)+30(24)=816.

Hence

12T+720=816,T=8.12T+720=816, \qquad\boxed{T=8}.

(c)

解法一

思路

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总时间为 3838 秒,货车匀速运动的时间是 38−12=2638-12=26 秒。用同样的面积方法求最高速率。

答题过程

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Both vehicles reach BB at time T+30=38 sT+30=38\text{ s}. For the van,

12(12)V+(38−12)V=816.\frac12(12)V+(38-12)V=816.

Thus

6V+26V=816,V=25.5.6V+26V=816, \qquad\boxed{V=25.5}.