📚 A-Level Physics Unit 3 Jan 22: Formula Derivation | A-Level 物理 Unit 3 2022年1月试卷 公式推导
In A-Level Physics Unit 3, especially the January 2022 examination paper, students are frequently asked to derive a linear relationship from a physics law, use experimental data to plot a graph, and then extract a physical constant such as the acceleration of free fall, resistivity, or a spring constant. These tasks combine practical skills with algebraic manipulation. One classic example is the determination of gravitational acceleration g using a free-fall experiment. This article unpacks the step‑by‑step derivation of the working formula s = ½ g t², transforms it into the graph-ready equation t² = (2/g)s, and explores how uncertainties propagate into the final result. Every step is explained with a focus on what examiners expect in Unit 3 January 2022‑style questions.
在 A-Level 物理 Unit 3 考试中，特别是 2022 年 1 月的试卷，经常要求学生从一个物理定律出发推导出线性关系，再利用实验数据绘图，最后求出诸如自由落体加速度、电阻率或弹簧常数这样的物理常量。这类任务把实验技能和代数处理结合在一起。一个经典的例子就是用自由落体实验测定重力加速度 g。本文会逐步拆解工作公式 s = ½ g t² 的推导过程，把它转换成适合作图的方程 t² = (2/g)s，并探讨不确定度是如何传递到最终结果中的。每一步都紧扣 Unit 3 2022 年 1 月考题的评分要求进行讲解。
1. Overview of Unit 3 Practical Skills | Unit 3 实验技能概览
Unit 3 of the A-Level Physics specification, whether from AQA, Edexcel or OCR, focuses on planning, implementing, analysing and evaluating practical work. The January 2022 question paper typically includes a scenario where a student carries out an experiment, records measurements, and must process the data. Deriving a suitable formula is often the first step because it determines which quantities should be plotted on the x‑ and y‑axes to produce a straight line. The gradient and intercept of that line then yield the target constant. Understanding the derivation is therefore not just a mathematical exercise – it is the foundation that links the physical law to the experimental method.
无论 AQA、Edexcel 还是 OCR 的 A-Level 物理大纲，Unit 3 的重点都是实验方案设计、实施、分析和评估。2022 年 1 月的试卷通常会给出一位学生进行实验、记录数据的情境，并要求考生处理这些数据。推导合适的公式往往是第一步，因为它决定了应该在 x 轴和 y 轴上画什么量才能得到一条直线。直线的斜率和截距随后就给出了目标常量。因此，理解推导过程不仅仅是数学练习，它更是把物理定律与实验方法连接起来的基础。
2. The Physics Behind Free Fall | 自由落体背后的物理
An object falling freely under gravity, assuming negligible air resistance, accelerates uniformly with acceleration g ≈ 9.81 m s⁻². If the object is released from rest, its initial velocity u = 0. The displacement s after a time t is given by the kinematic equation: s = u t + ½ a t². Substituting u = 0 and a = g yields s = ½ g t². This equation tells us that the distance fallen is directly proportional to the square of the time – a non‑linear relationship. To obtain a straight‑line graph, we need to rearrange the equation into the form y = m x + c.
在忽略空气阻力的情况下，物体仅在重力作用下自由下落时做匀加速运动，加速度 g 约等于 9.81 m s⁻²。如果物体从静止开始释放，其初速度 u = 0。经过时间 t 后的位移 s 由运动学方程 s = u t + ½ a t² 给出。代入 u = 0 和 a = g 便得到 s = ½ g t²。这个方程告诉我们，下落距离正比于时间的平方——这是一个非线性关系。要得到直线图线，我们需要把方程重新整理成 y = m x + c 的形式。
3. Deriving the Straight‑Line Equation | 导出直线方程
Starting with s = ½ g t², we can divide both sides by ½ g to isolate t², but a more examiner‑friendly approach is to treat t² as the dependent variable. Multiply both sides by 2: 2s = g t². Then divide by g: t² = (2/g) s. Now the equation is in the form y = m x, where y ≡ t², x ≡ s, and the gradient m = 2/g. There is no intercept because the line passes through the origin (when s = 0, t² = 0). This is exactly what a Unit 3 question expects you to recognise: plotting t² on the vertical axis and s on the horizontal axis should give a straight line through the origin, and the gradient equals 2/g.
从 s = ½ g t² 出发，我们可以两边除以 ½ g，把 t² 单独解出来，但更符合评分习惯的做法是把 t² 看作因变量。两边乘以 2 得到 2s = g t²，然后除以 g，得到 t² = (2/g) s。现在方程就是 y = m x 的形式，其中 y ≡ t²，x ≡ s，斜率 m = 2/g。没有截距，因为图线过原点（当 s = 0 时，t² = 0）。这正是 Unit 3 考题希望你看出来的：纵轴画 t²，横轴画 s，应当得到一条过原点的直线，且斜率等于 2/g。
4. Graphical Analysis in Unit 3 Jan 22 | Unit 3 2022 年 1 月试题中的图线分析
The January 2022 paper might present a table of s and t values, each measured with an uncertainty. The candidate is required to calculate t² for each reading, plot a graph of t² against s, draw a line of best fit, and determine the gradient. From m = 2/g, we can rearrange to find g = 2/m. If the best‑fit line gives, for example, m = 0.203 s² m⁻¹, then g = 2 / 0.203 ≈ 9.86 m s⁻². The percentage uncertainty in g is directly linked to the uncertainty in the gradient, which can be found by drawing worst‑fit lines. The derivation of the formula is the logical thread that holds the entire analysis together.
2022 年 1 月的试卷可能会给出一张包含 s 和 t 的数据表，每个量都带有不确定度。考生需要为每个读数计算 t²，画出 t²‑s 图，画一条最佳拟合线，并求出斜率。由 m = 2/g，整理可得 g = 2/m。如果最佳拟合线给出斜率 m = 0.203 s² m⁻¹，那么 g = 2 / 0.203 ≈ 9.86 m s⁻²。g 的百分不确定度直接和斜率的不确定度相关，后者可以通过画最差拟合线得到。公式推导就像一条逻辑线，把整个分析串在一起。
5. Step‑by‑Step Algebraic Manipulation | 代数处理的步骤分解
Let us write the derivation clearly so that no marks are lost in an exam. The given physical law is s = ½ g t². Step 1: Multiply both sides by 2 → 2s = g t². Step 2: Divide both sides by g → t² = (2/g) s. Step 3: Identify the linear form → y = m x + c with y = t², x = s, m = 2/g, c = 0. Always state that a graph of t² against s is expected to be a straight line through the origin. In Unit 3, marks are awarded for the explicit linking of the equation to the graph. Do not skip the step of stating that c = 0; otherwise, a non‑zero intercept could be misinterpreted as a systematic error.
让我们把推导过程清晰地写出来，确保考试中不失分。已知物理定律是 s = ½ g t²。步骤 1：两边同乘以 2 → 2s = g t²。步骤 2：两边同除以 g → t² = (2/g) s。步骤 3：识别线性形式 → y = m x + c，其中 y = t²，x = s，m = 2/g，c = 0。一定要写出，预期 t²‑s 图是一条过原点的直线。在 Unit 3 中，明确把方程和图线联系起来是可以得分的。不要省略说明 c = 0 这一步；否则非零截距可能会被错误地当成系统误差。
6. Common Mistakes in Deriving the Formula | 公式推导中的常见错误
A frequent error is plotting t against s instead of t² against s. The raw relationship s ∝ t² is a parabola, not a straight line, and exam questions often test whether students can linearise it. Another mistake is forgetting the factor of ½ or misplacing g. Some students write t² = g s / 2, which confuses the gradient. Be methodical: after rearranging, check dimensions. The left side t² has units of s²; the right side (2/g)s must also have units of s². Since g is in m s⁻², 1/g is s² m⁻¹, multiplied by s (metres) gives s², confirming the derivation is dimensionally consistent.
一个常见的错误是画 t‑s 图，而不是 t²‑s 图。原始关系 s ∝ t² 是抛物线，不是直线，考题经常考察学生是否会作线性化处理。另一个错误是忘记因子 ½ 或者把 g 放错位置。有些学生会写成 t² = g s / 2，这就把斜率弄混了。推导时要条理分明：整理之后检查量纲。等号左边 t² 的单位是 s²；右边 (2/g)s 也必须是 s²。因为 g 的单位是 m s⁻²，1/g 是 s² m⁻¹，乘以 s（米）得到 s²，这就证实了推导在量纲上是一致的。
7. Including Uncertainties in the Derived Formula | 在导出公式中考虑不确定度
Unit 3 papers place a strong emphasis on measurement uncertainties. When we derive t² = (2/g) s, we must also consider how the uncertainty in each measured quantity affects the final value of g. Typically, the uncertainty in the gradient Δm is found using the difference between the best‑fit and worst‑fit slopes. Because g = 2/m, the percentage uncertainty in g equals the percentage uncertainty in m: %U(g) = %U(m). This is a direct consequence of the formula. Occasionally, if the intercept is not exactly zero, the question may ask you to derive a modified formula that includes an intercept term, e.g. t² = (2/g)s + c, and discuss its physical meaning (such as reaction time).
Unit 3 试卷非常注重测量不确定度。当我们推导出 t² = (2/g) s 时，也必须考虑每个测量量的不确定度是如何影响最终 g 值的。通常，斜率的不确定度 Δm 是通过最佳拟合线与最差拟合线斜率之差得到的。因为 g = 2/m，g 的百分不确定度就等于 m 的百分不确定度：%U(g) = %U(m)。这是由公式直接得出的结论。偶尔，如果截距不恰好为零，题目可能会要求你推导一个包含截距项的修正公式，例如 t² = (2/g)s + c，并讨论其物理意义（例如人的反应时间）。
8. Worked Example from a Jan‑22 Style Question | 一道 Jan‑22 风格例题的完整推演
Imagine a typical Unit 3 item: a student drops a ball‑bearing from rest and uses a trapdoor and electronic timer to measure the time of fall for various heights s. The data are: s = 0.200 m, t = 0.202 s; s = 0.400 m, t = 0.286 s; s = 0.600 m, t = 0.350 s; s = 0.800 m, t = 0.404 s; s = 1.000 m, t = 0.452 s. Calculate t² for each: 0.0408, 0.0818, 0.1225, 0.1632, 0.2043 s². Plot the graph and find the gradient. Suppose m = 0.205 s² m⁻¹. Using the derived formula g = 2/m, we obtain g = 2 / 0.205 = 9.76 m s⁻². The derivation links the raw data to the final answer in a transparent, exam‑ready chain of logic.
设想一道典型的 Unit 3 题目：一位学生从静止释放小球，用活动门和电子计时器测量不同高度 s 的下落时间。数据为：s = 0.200 m，t = 0.202 s；s = 0.400 m，t = 0.286 s；s = 0.600 m，t = 0.350 s；s = 0.800 m，t = 0.404 s；s = 1.000 m，t = 0.452 s。计算每个 t²：0.0408、0.0818、0.1225、0.1632、0.2043 s²。画图并求出斜率。假设 m = 0.205 s² m⁻¹。利用推导出的公式 g = 2/m，得到 g = 2 / 0.205 = 9.76 m s⁻²。整个推导把原始数据与最终答案连成了一条清晰的、符合考试要求的逻辑链。
9. Extending the Idea to Other Unit 3 Experiments | 将推导思路扩展到其他 Unit 3 实验
The principle of deriving a linear formula is not limited to free fall. In the same January 22 paper, you might encounter a second experiment, such as determining the resistivity of a metal wire. From R = ρ L / A, the linearised form is R = (ρ/A) L. Plotting R against L gives a gradient of ρ/A, allowing ρ to be calculated. Another common scenario is a spring‑mass system, where T = 2π √(m/k) is squared to give T² = (4π²/k) m. Recognising the underlying pattern – identify the law, isolate the variable that gives a straight line, and interpret the slope – is a transferable skill that scores highly in Unit 3.
导出线性公式的原理不限于自由落体。在同样的 2022 年 1 月试卷中，你可能会遇到第二个实验，比如测定金属丝的电阻率。由 R = ρ L / A，线性化后得到 R = (ρ/A) L。画出 R‑L 图，斜率就是 ρ/A，从而可以算出 ρ。另一个常见情景是弹簧‑质量系统，将 T = 2π √(m/k) 两边平方得到 T² = (4π²/k) m。看出底层模式——确定定律，分离变量得到直线，解释斜率——是一种可迁移的技能，在 Unit 3 中能拿高分。
10. Final Checklist for the Derivation Section | 推导部分的最终检查清单
To secure full marks on the formula‑derivation task in Unit 3 Jan 22, ensure you: (1) write the fundamental physical law correctly; (2) rearrange it stepwise, showing all algebraic moves; (3) state the quantities to be plotted on each axis; (4) express the gradient in terms of the constant you are trying to find; (5) remark that the line should pass through the origin, and state why (c = 0). Wherever possible, support your derivation with a dimension check. This thoroughness not only satisfies the mark scheme but also reduces careless errors.
要在 Unit 3 2022 年 1 月试卷的公式推导任务中拿到满分，请确保做到以下几点：(1) 正确写出基本物理定律；(2) 逐步整理方程，展示所有代数过程；(3) 说明要在两轴上画的量；(4) 用待求的常量表达斜率；(5) 指出图线应当过原点，并说明原因（c = 0）。只要有可能，就用量纲检查来支持你的推导。这种细致不仅能满足评分标准，还能减少粗心导致的错误。
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