📚 Deriving Kinematic Equations from AS Physics Unit 1 Jan 2019 | 从2019年1月AS物理真题推导运动学公式
In AS Physics Unit 1, students frequently encounter questions that require them to derive the equations of uniformly accelerated motion from fundamental definitions and graphical analysis. This article focuses on the derivations commonly examined in past papers, particularly referencing the style of January 2019. By working through the velocity-time graph approach, you will not only memorise the equations but also understand their physical meaning, which is essential for tackling proof-style questions and explaining the steps clearly.
在AS物理第一单元中，学生常会遇到要求从基本定义和图形分析推导匀加速运动方程的题目。本文重点讲解历年真题中常见的推导过程，特别参考2019年1月试卷的风格。通过速度-时间图的推导方法，你不仅能记住公式，还能理解其物理意义，这对解答证明类题目和清晰解释步骤至关重要。
1. Understanding Uniform Acceleration | 理解匀加速运动
Uniform acceleration occurs when an object’s velocity changes by the same amount in each equal time interval. In this scenario, the acceleration a is constant. The standard symbols used are u for initial velocity, v for final velocity, t for time, and s for displacement. All vector directions are typically simplified to motion along a straight line, so positive and negative signs indicate direction.
匀加速运动发生时，物体在每段相等的时间内速度变化量相同。在此情况下，加速度a恒定。标准符号为：初速度u、末速度v、时间t和位移s。所有矢量方向通常简化为沿直线运动，因此正负号表示方向。
When tackling a derivation question, the examiner is testing your ability to connect the definitions of acceleration and average velocity to geometric representations. You must be able to start from a = (v – u)/t and the idea that displacement equals the area under a velocity-time graph. These two starting points form the backbone of all subsequent kinematic derivations.
在解答推导题时，考官考察的是你将加速度和平均速度的定义与几何表示联系起来的能力。你必须能够从 a = (v – u)/t 以及位移等于速度-时间图下面积这一概念入手。这两个出发点构成了所有后续运动学推导的基石。
2. The Velocity-Time Graph | 速度-时间图
For uniform acceleration, the velocity-time graph is a straight line with a constant gradient equal to the acceleration a. The line starts at (0, u) and ends at (t, v). The area beneath this line represents the displacement s. This graphical interpretation is crucial because it allows us to express s in terms of the average velocity and time without any integration.
对于匀加速运动，速度-时间图是一条直线，其恒定斜率等于加速度a。该直线从点(0, u)出发，终止于点(t, v)。直线下方的面积代表位移s。这种图形解释至关重要，因为它使我们能够用平均速度和时间表示位移，而无需积分。
In the Jan 2019 series, many mark schemes rewarded a clear sketch or description of the v-t graph before writing the equations. You can simply state that the area is a trapezium (trapezoid). The parallel sides correspond to u and v, and the height is t. The area formula for a trapezium gives a quick route to one of the fundamental motion equations.
在2019年1月的考试系列中，许多评分方案奖励学生在写出方程前先绘制或描述v-t图的清晰做法。你可以直接说明该区域是一个梯形。平行边对应u和v，高为t。梯形面积公式为得到基本运动方程之一提供了快捷途径。
3. Deriving s = ½(u+v)t | 推导 s = ½(u+v)t
From the definition of average velocity, displacement s is the average velocity multiplied by time. For uniform acceleration, the average velocity is exactly ½(u+v), since the velocity changes linearly. Therefore, we can write the first core equation without using acceleration:
根据平均速度的定义，位移s等于平均速度乘以时间。对于匀加速运动，由于速度线性变化，平均速度恰好为½(u+v)。因此，我们可以在不使用加速度的情况下写出第一个核心方程：
s = ½(u+v)t
This equation is also directly obtained from the area under the v-t graph. The trapezium area is ½ × (sum of parallel sides) × height = ½(u+v)t. Many past paper questions, including those in the Jan 2019 paper, ask you to show this step explicitly. Always state that the area under the line equals the displacement.
该方程也可直接从v-t图下的面积得到。梯形面积等于½ ×（平行边之和）× 高 = ½(u+v)t。许多历年真题，包括2019年1月的试卷，都要求明确展示这一步骤。一定要说明直线下方的面积等于位移。
4. Deriving v = u + at | 推导 v = u + at
The definition of acceleration is the rate of change of velocity. Mathematically, a = (v – u)/t. This is the gradient of the velocity-time graph. By rearranging this definition, we obtain the first of the standard ‘suvat’ equations that explicitly includes acceleration:
加速度的定义是速度的变化率。数学上表示为 a = (v – u)/t。这正是速度-时间图的斜率。通过重新整理这个定义，我们得到第一个明确包含加速度的标准“suvat”方程：
v = u + at
This derivation is often the easiest, but students must be careful to present it as stemming from the definition a = Δv/Δt, not merely stating the final equation. In a Jan 2019 style question, you might be instructed to ‘State the relationship between acceleration, velocity and time. Hence derive v = u + at.’ You should write the defining equation and then multiply both sides by t and add u.
这个推导通常最简单，但学生必须注意，要表明它源自定义 a = Δv/Δt，而不仅仅是写出最终方程。在2019年1月风格的题目中，可能会要求“陈述加速度、速度和时间的关系，并由此推导 v = u + at”。你应该写出定义式，然后两边乘以 t 并加上 u。
5. Deriving s = ut + ½at² | 推导 s = ut + ½at²
To eliminate the final velocity v from the displacement equation, we substitute v = u + at into s = ½(u+v)t. This substitution is a common requirement for 3-4 mark proof questions. Begin by writing s = ½(u + (u + at))t, then simplify inside the brackets to get (2u + at).
为了从位移方程中消去末速度v，我们将 v = u + at 代入 s = ½(u+v)t。这种代入是3-4分证明题的常见要求。首先写出 s = ½(u + (u + at))t，然后化简括号内为 (2u + at)。
Next, multiply through by the ½ and the t: s = ½(2ut + at²) = ut + ½at². Each algebraic step should be shown clearly. Many mark schemes for the Jan 2019 paper award marks for intermediate expansion and correct handling of the factor ½.
接着，将½和t乘入：s = ½(2ut + at²) = ut + ½at²。每一步代数步骤都应清晰展示。2019年1月试卷的许多评分方案会因中间展开步骤和正确处理½因子而给分。
s = ut + ½at²
This is the form used when the final velocity is unknown but acceleration and time are given. It also reveals the displacement as the sum of the distance covered due to initial velocity and the additional distance from acceleration.
这是当末速度未知但加速度和时间已知时使用的形式。它还将位移揭示为因初速度覆盖的距离与加速度产生的附加距离之和。
6. Deriving v² = u² + 2as | 推导 v² = u² + 2as
When time t is not needed, we eliminate t between v = u + at and s = ½(u+v)t. One approach is to rewrite v = u + at as t = (v – u)/a. Then substitute this into s = ½(u+v) × (v – u)/a. Recognise that (u+v)(v-u) = v² – u².
当不需要时间t时，我们联立 v = u + at 和 s = ½(u+v)t 消去t。一种方法是把 v = u + at 改写为 t = (v – u)/a，然后代入 s = ½(u+v) × (v – u)/a。注意到 (u+v)(v-u) = v² – u²。
Thus, s = ½ × (v² – u²)/a. Multiplying both sides by 2a yields the well-known equation. Make sure to state that this equation is useful for problems involving speed and distance without time.
因此，s = ½ × (v² – u²)/a。两边乘以2a即得到广为人知的方程。务必说明该方程适用于不涉及时间的速度和距离问题。
v² = u² + 2as
Alternatively, you can start from s = ut + ½at² and v = u + at, then square v and compare. The Jan 2019 paper often awards full marks if the derivation begins with the two basic equations and shows the algebraic manipulation clearly.
或者，你可以从 s = ut + ½at² 和 v = u + at 出发，然后将 v 平方并比较。2019年1月的试卷通常会在推导从两个基本方程开始并清晰展示代数运算时给予满分。
7. Applying Derivations to Past Paper Questions | 将推导应用于真题
A typical question from AS Unit 1 Jan 2019 might ask: ‘A car accelerates uniformly from 8.0 m s⁻¹ to 20 m s⁻¹ over a distance of 84 m. By deriving the appropriate SUVAT equation, show that its acceleration is 2.0 m s⁻².’ You would select v² = u² + 2as, write it down, and state that it comes from equating definitions and substituting for t.
AS第一单元2019年1月的一道典型题目可能会问：“一辆汽车从8.0 m s⁻¹匀加速到20 m s⁻¹，经过84 m。通过推导适当的SUVAT方程，证明其加速度为2.0 m s⁻²。”你会选择 v² = u² + 2as，写下它，并说明它来自联立定义并代入t。
Other questions ask for the derivation itself without numbers. For instance, ‘Using the graph of v against t, show that s = ut + ½at².’ Here you must refer to the area of a trapezium and the equation of the straight line v = u + at. Present your answer step by step: area = s, gradient = a, then substitute.
其他问题则要求不含数字的推导本身。例如，“利用v与t的关系图，证明 s = ut + ½at²。”此时你必须提及梯形面积和直线方程 v = u + at。逐步展示你的答案：面积 = s，斜率 = a，然后代入。
Practising these proof-style answers from past papers ensures you can reproduce the logic under exam conditions. Always label your starting equations and states your assumptions (constant acceleration, straight line motion).
从历年真题中练习这类证明型答案，能确保你在考试条件下重现该逻辑。务必标注起始方程并说明你的假设（恒定加速度，直线运动）。
8. Common Mistakes and Tips | 常见错误与技巧
One frequent error is confusing the average velocity for uniform acceleration with that for constant velocity. Remember that (u+v)/2 applies only when acceleration is constant. Another mistake is forgetting to square the units when substituting values, but in derivation questions this is less relevant as you are working symbolically.
一个常见错误是将匀加速运动的平均速度与匀速运动的平均速度混淆。记住 (u+v)/2 仅在加速度恒定时适用。另一个错误是代入数值时忘记对单位平方，但在推导题中这一点不那么相关，因为你是进行符号运算。
Many students lose marks by omitting the step that connects area to displacement. Always explicitly write ‘Displacement = area under v-t graph’ before using the trapezium area formula. Also, when deriving v² = u² + 2as, do not merely write the final equation; the mark scheme expects to see t eliminated.
许多学生因遗漏将面积与位移联系起来的步骤而丢分。在使用梯形面积公式之前，务必明确写出“位移 = v-t图下面积”。此外，在推导 v² = u² + 2as 时，不要仅仅写出最终方程；评分方案期望看到 t 被消去的过程。
For top marks, present your derivation as a logical sequence. Start from fundamental definitions, sketch the graph if it helps, and show algebraic rearrangement neatly. The Jan 2019 examiners’ report highlighted that students who wrote clear sub-steps were rewarded even if a minor algebraic slip occurred later.
要获得高分，将你的推导呈现为一个逻辑序列。从基本定义出发，如有帮助可绘制草图，并整洁地展示代数变形。2019年1月的考官报告强调，即使后来出现细微代数失误，写出清晰子步骤的学生仍能获得分数。
9. Summary Table of Equations | 方程总结表
The following table lists the four essential equations for uniformly accelerated motion, along with the quantities each equation relates. Knowing when each equation is useful will speed up your exam responses.
下表列出了匀加速运动的四个基本方程，以及每个方程关联的量。了解何时使用每个方程将加快你的考试答题速度。
| Equation | Variables Involved | Missing Quantity |
|---|---|---|
| v = u + at | v, u, a, t | s |
| s = ½(u+v)t | s, u, v, t | a |
| s = ut + ½at² | s, u, t, a | v |
| v² = u² + 2as | v, u, a, s | t |
Practise deriving these equations in different orders until you can do it from memory. This will give you confidence when facing the proof-style question that is almost certainty to appear in your AS Unit 1 exam, just as it did in Jan 2019.
练习以不同次序推导这些方程，直到你能凭记忆完成。这将使你在面对几乎肯定会在AS第一单元考试中出现的证明型题目时充满信心，就像2019年1月那样。
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