Essential Formula Derivations for OxfordAQA International AS & A-Level Physics | OxfordAQA 国际 AS 与 A-Level 物理核心公式推导

📚 Essential Formula Derivations for OxfordAQA International AS & A-Level Physics | OxfordAQA 国际 AS 与 A-Level 物理核心公式推导

Understanding the derivation of key equations in the OxfordAQA International A-Level Physics specification not only strengthens your conceptual grasp but also prepares you for higher-order exam questions. This article walks through the essential derivations, from kinematics to fields and kinetic theory, with clear bilingual explanations paired step by step.

掌握 OxfordAQA 国际 A-Level 物理大纲中关键公式的推导过程，不仅能加深概念理解，还能帮你从容应对高阶考题。本文以清晰的双语解释，逐步梳理从运动学到场与分子动理论的核心推导。

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1. Deriving the SUVAT Equations | 匀变速运动方程推导

For constant acceleration a, the definition a = (v − u)/t immediately gives the first equation.

v = u + a t

加速度恒定时，由定义式 a = (v − u)/t 直接移项即可得到第一个运动学方程。

Average velocity during uniform acceleration is (u + v)/2, so displacement s equals average velocity multiplied by time.

s = ½ (u + v) t

匀加速运动的平均速度为 (u + v)/2，因此位移 s = 平均速度 × 时间。

Substituting v = u + at into the displacement formula eliminates v, yielding the standard form.

s = u t + ½ a t²

将第一个式子代入消去 v，整理得到含时间的位移公式。

Finally, eliminating time t between v = u + at and s = ½(u + v)t gives the timeless relation.

v² = u² + 2 a s

最后，从两式中消去时间 t，得到不含时间的速度-位移关系。

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2. Projectile Motion Trajectory | 抛体运动轨迹方程

Resolve the launch velocity u at angle θ: horizontal component u cosθ, vertical component u sinθ. Horizontally, x = (u cosθ) t; vertically, with a = −g, y = (u sinθ) t − ½ g t². Eliminate t to obtain the parabolic trajectory.

y = x tan θ − (g x²) / (2 u² cos² θ)

将初速度分解为水平 u cosθ 和竖直 u sinθ。水平方向匀速，x = u cosθ·t；竖直方向加速度 −g，y = u sinθ·t − ½ g t²。消去时间 t，即得抛物线轨迹方程。

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3. Centripetal Acceleration a = v²/r | 向心加速度公式推导

An object moving at constant speed v in a circle of radius r experiences a changing velocity direction. In a short time Δt, the velocity change Δv points toward the centre. By similar triangles, Δv/v = Δs/r = vΔt/r, so the magnitude of acceleration is a = Δv/Δt = v²/r.

a = v² / r = ω² r

物体以恒定速率 v 做半径为 r 的圆周运动，速度方向持续变化。极短时间内速度变化量 Δv 指向圆心，由几何相似性得 Δv/v = Δs/r = v·Δt/r，因此向心加速度大小 a = Δv/Δt = v²/r，也可写作 ω² r。

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4. Simple Harmonic Motion & Angular Frequency | 简谐运动与角频率

In SHM, restoring force F = −k x leads to acceleration a = F/m = −(k/m) x. Defining ω² = k/m gives the fundamental equation of SHM: a = −ω² x. The solution is x = A cos(ωt + φ), and the period follows as T = 2π/ω.

a = −ω² x , T = 2π √(m/k)

简谐运动中回复力 F = −k x，加速度 a = −(k/m)x。令 ω² = k/m，则 a = −ω² x，这是简谐运动的特征方程。其解为 x = A cos(ωt + φ)，周期 T = 2π/ω = 2π√(m/k)。

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5. Period of a Simple Pendulum | 单摆周期公式推导

For a small angular displacement, the restoring force along the arc is approximately −mg(x/L), where x is the arc length. Hence a = −(g/L) x. Comparing with a = −ω² x yields ω² = g/L, so the period is T = 2π √(L/g).

T = 2π √(L / g)

在小角度摆动下，回复力约等于 −mg(x/L)，加速度 a = −(g/L)x。与简谐运动标准形式对比，得 ω² = g/L，因此周期 T = 2π√(L/g)。该公式仅在小角度近似下成立。

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6. Gravitational Field Strength g = GM/r² | 引力场强度推导

Newton’s law of gravitation gives F = G M m / r². Since gravitational field strength g is defined as force per unit mass, g = F/m, we directly obtain g = GM / r².

g = G M / r²

由牛顿万有引力定律 F = GMm/r²，引力场强度 g 定义为单位质量所受的引力，代入即得径向场强公式 g = GM/r²。该式适用于质点外或球对称质量外部的场。

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7. Electric Field Strength Derivations | 电场强度公式推导

From Coulomb’s law, the force between point charges is F = k Q q / r². Electric field strength E = F/q, so for a point charge E = k Q / r². For a uniform field between parallel plates, potential difference V is related to work done: qV = F d = q E d, thus V = E d.

E = k Q / r²

E = V / d

根据库仑定律 F = kQq/r²，电场强度 E = F/q，点电荷的场强即为 E = kQ/r²。在平行板匀强电场中，电势差 V 与电场力做功的关系为 qV = qEd，因此 V = Ed，即 E = V/d。

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8. Capacitor Discharge Equation | 电容放电方程推导

During discharge through a resistor R, the current I equals the rate of decrease of charge: I = −dQ/dt. Using Ohm’s law V = IR and the definition C = Q/V gives V = Q/C. Equating: Q/C = −R dQ/dt, which rearranges to dQ/dt = −Q/(RC). Solving this first-order differential equation gives an exponential decay.

Q = Q₀ e^−t/(RC)

电容通过电阻放电时，电流 I = −dQ/dt。由欧姆定律 V = IR，且 C = Q/V，得 V = Q/C。联立有 Q/C = −R dQ/dt，即 dQ/dt = −Q/(RC)。求解此一阶微分方程，得到电荷随时间指数

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