📚 A-Level Physics: Formula Derivations from June 2018 Question Paper 5 | A-Level 物理：2018年6月卷5公式推导
In the June 2018 A Level Physics Paper 5, students were required not only to apply physical formulas but also to understand their origins. Mastering the derivations of key equations strengthens conceptual understanding and prepares you for both experimental planning and theoretical questions. This article revisits the essential derivations linked to that paper, breaking them down step by step.
在2018年6月的A Level物理卷5中，学生不仅需要应用物理公式，还要理解它们的来源。掌握关键方程的推导能强化概念理解，为实验设计和理论题做好准备。本文重温与那张试卷相关的基本推导，逐步拆解每一个公式的由来。
1. Understanding Physics Derivations in Paper 5 | 理解卷5中的物理推导
Paper 5 often tests the ability to derive expressions from first principles, linking theory to practical scenarios. Whether it is kinematics, circular motion, or energy storage, a logical sequence of algebraic steps is expected.
卷5经常考查从基本原理推导表达式的能力，将理论与实际情境相联系。无论是运动学、圆周运动还是能量储存，都需要清晰的代数推导步骤。
These derivations are not isolated exercises – they form the foundation of the data analysis and evaluation tasks in the paper. Being fluent in them saves time and reduces errors when manipulating experimental relationships.
这些推导并非孤立的练习——它们构成了试卷中数据分析和评估任务的基础。熟练掌握这些推导能在处理实验关系时节省时间并减少错误。
2. Deriving v = u + at from Acceleration Definition | 从加速度定义推导 v = u + at
Acceleration is defined as the rate of change of velocity. For constant acceleration a over a time interval t, we can write:
加速度定义为速度的变化率。对于在时间间隔 t 内的恒定加速度 a，我们可以写出：
a = (v – u) / t
Multiplying both sides by t gives at = v – u. Adding u to both sides yields the first SUVAT equation.
两边乘以 t 得到 at = v – u。两边加上 u 就得到第一条 SUVAT 方程。
v = u + at
This derivation assumes the object moves in a straight line with uniform acceleration. It is the starting point for all other kinematic equations.
这一推导假设物体以匀加速度沿直线运动。它是所有其他运动学方程的起点。
3. Deriving s = ut + ½ at² using Average Velocity | 利用平均速度推导 s = ut + ½ at²
For constant acceleration, the average velocity vav is the mean of the initial and final velocities:
对于匀加速度，平均速度 vav 是初速度和末速度的平均值：
vav = (u + v) / 2
Displacement s is average velocity multiplied by time: s = vav t. Substituting the expression for vav and then replacing v with u + at from the first equation gives:
位移 s 等于平均速度乘以时间：s = vav t。代入 vav 的表达式，再用第一条方程中的 v = u + at 替换 v，得到：
s = ((u + u + at) / 2) × t = (2u + at) t / 2
Simplifying leads directly to the displacement equation with the familiar ½ at² term.
化简后直接得到含有熟悉的 ½ at² 项的位移方程。
s = ut + ½ at²
4. Deriving v² = u² + 2as by Eliminating t | 消去时间推导 v² = u² + 2as
To eliminate time t, rearrange the equation v = u + at into t = (v – u)/a. Then substitute this t into s = ut + ½ at².
为了消去时间 t，将方程 v = u + at 变形为 t = (v – u)/a。然后将此 t 代入 s = ut + ½ at²。
s = u((v – u)/a) + ½ a ((v – u)/a)²
Expanding and simplifying the terms produces (uv – u²)/a + (v² – 2uv + u²)/(2a). Combine into a single fraction and multiply by 2a.
展开并化简各项得出 (uv – u²)/a + (v² – 2uv + u²)/(2a)。合并为单个分式并两边同乘 2a。
2as = 2uv – 2u² + v² – 2uv + u² = v² – u²
Adding u² to both sides gives the time-independent kinematic relation.
两边加上 u² 便得到与时间无关的运动学关系。
v² = u² + 2as
5. Graphical Derivation of Displacement for Uniform Acceleration | 匀加速位移的图形推导
On a velocity–time graph, constant acceleration appears as a straight line starting at u and ending at v. The area under the graph represents displacement.
在速度–时间图上，匀加速度表现为一条由 u 开始、到 v 结束的直线。图线下方的面积代表位移。
Splitting the area into a rectangle of height u and a triangle of height (v – u) gives: s = ut + ½ (v – u)t. Using v – u = at confirms s = ut + ½ at².
将面积拆分为高度为 u 的矩形和高度为 (v – u) 的三角形，得到：s = ut + ½ (v – u)t。利用 v – u = at 即可确认 s = ut + ½ at²。
This geometric approach reinforces why the ½ factor appears and is particularly useful when analysing experimental motion data in Paper 5.
这种几何方法强化了为什么会出现 ½ 因子，在分析卷5中的实验运动数据时特别有用。
6. Deriving Centripetal Acceleration a = v²/r | 向心加速度公式 a = v²/r 推导
Consider an object moving at constant speed v in a circle of radius r. Over a small time Δt, the direction of velocity changes by Δθ, but its magnitude remains v.
考虑一个物体以恒定速率 v 在半径为 r 的圆周上运动。在很短的时间 Δt 内，速度方向改变了 Δθ，但其大小保持为 v。
The velocity vector triangle is similar to the displacement triangle formed by the two radii. Therefore, Δv / v = chord length / r. For small Δθ, the chord length approximates the arc length vΔt.
速度矢量三角形与两条半径构成的位移三角形相似。因此，Δv / v = 弦长 / r。对于小角度 Δθ，弦长近似等于弧长 vΔt。
Δv / v = (v Δt) / r
Acceleration is Δv/Δt, so a = v²/r. The direction of this acceleration is towards the centre of the circle.
加速度等于 Δv/Δt，因此 a = v²/r。该加速度的方向指向圆心。
Using the relation v = ωr, the centripetal acceleration is also written as a = ω²r, which is essential for circular motion questions in Paper 5.
利用关系式 v = ωr，向心加速度也可写成 a = ω²r，这对卷5中的圆周运动问题至关重要。
7. Deriving Simple Harmonic Motion Acceleration a = – ω²x | 简谐运动加速度 a = – ω²x 推导
Simple harmonic motion (SHM) can be modelled as the projection of uniform circular motion onto a diameter. If a point moves in a circle of radius A with angular speed ω, its displacement from the centre in the x-direction is x = A cos(ωt).
简谐运动 (SHM) 可以看作是匀速圆周运动在某条直径上的投影。如果一个点以角速度 ω 在半径为 A 的圆上运动，它在 x 方向上的位移为 x = A cos(ωt)。
Velocity is the time derivative of displacement: v = –Aω sin(ωt). Acceleration is the derivative of velocity: a = –Aω² cos(ωt).
速度是位移的时间导数：v = –Aω sin(ωt)。加速度是速度的导数：a = –Aω² cos(ωt)。
Since A cos(ωt) = x, we obtain a = –ω²x. The negative sign indicates that acceleration is always directed towards the equilibrium position.
由于 A cos(ωt) = x，我们得到 a = –ω²x。负号表示加速度始终指向平衡位置。
This defining equation of SHM is frequently the starting point for oscillation analysis in practical Paper 5 tasks, such as investigating a mass-spring system.
这个 SHM 的定义方程常常是卷5实验任务中分析振动的起点，例如研究弹簧质量系统。
8. Deriving Period of a Mass-Spring System T = 2π√(m/k) | 弹簧振子周期推导
For a mass m attached to a spring of stiffness k, Hooke’s law gives the restoring force F = – kx. Newton’s second law states F = ma, thus ma = – kx.
对于连接在劲度系数为 k 的弹簧上的质量 m，胡克定律给出回复力 F = – kx。牛顿第二定律指出 F = ma，因此 ma = – kx。
a = – (k/m) x
Comparing this with the SHM equation a = – ω²x shows that ω² = k/m. Angular frequency ω is related to period T by ω = 2π/T.
将此式与简谐运动方程 a = – ω²x 比较，可知 ω² = k/m。角频率 ω 与周期 T 的关系为 ω = 2π/T。
(2π/T)² = k/m → T = 2π√(m/k)
This derivation is directly applicable when students plan experiments in Paper 5 to determine spring constants or unknown masses using oscillation timings.
当学生在卷5中设计实验，利用振动时间测定弹簧劲度系数或未知质量时，这一推导能直接应用。
9. Deriving Electrical Power Equations P = IV, I²R, V²/R | 电功率公式推导
Electric power P is defined as the energy transferred per unit time. When a charge Q moves through a potential difference V, the energy transferred is QV.
电功率 P 定义为单位时间内传递的能量。当电荷 Q 通过电势差 V 时，所传递的能量为 QV。
Current I is the rate of flow of charge, I = Q/t, so Q = It. Substituting gives P = VIt / t = IV.
电流 I 是电荷的流动速率，I = Q/t，因此 Q = It。代入后得到 P = VIt / t = IV。
P = I V
Using Ohm’s law V = IR, we can replace V or I to derive alternative forms: P = I(IR) = I²R, and P = (V/R)V = V²/R.
利用欧姆定律 V = IR，我们可以替换 V 或 I 来推导其他形式：P = I(IR) = I²R 以及 P = (V/R)V = V²/R。
These power relationships are vital when handling circuit analysis questions in the data section of Paper 5, where selecting the correct formula simplifies calculations.
在处理卷5数据部分的电路分析问题时，这些功率关系至关重要，选择合适的公式能简化计算。
10. Deriving Energy Stored in a Capacitor E = ½ CV² | 电容器储能推导
When charging a capacitor, the potential difference V across it builds up from 0 to the supply voltage. The charge stored is Q = CV.
给电容器充电时，其两端的电势差 V 从 0 逐渐升高至电源电压。储存的电荷量为 Q = CV。
Energy stored is the total work done moving each small charge element Δq through the instantaneous potential difference v. Summing all contributions gives:
储存的能量是将每一小份电荷 Δq 在瞬时电势差 v 下移动所做的总功。将所有贡献加总得到：
E = ∫ v dq
Using v = q/C and integrating from q=0 to Q, we get E = (1/C) ∫ q dq = (1/C) × ½ Q² = ½ Q²/C.
利用 v = q/C 并从 q=0 到 Q 积分，我们得到 E = (1/C) ∫ q dq = (1/C) × ½ Q² = ½ Q²/C。
Substituting Q = CV yields the well-known formula for capacitor energy in terms of capacitance and voltage.
代入 Q = CV 便得到用电容和电压表示的电容器能量公式。
E = ½ CV²
Understanding the derivation helps explain why half of the energy supplied by the battery is dissipated in the circuit resistance, a nuanced point occasionally probed in Paper 5 evaluations.
理解这一推导有助于解释为何电池提供的一半能量会在电路电阻中耗散，这是卷5评估中偶尔会探讨的细微之处。
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