📚 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。
1. Deriving the SUVAT Equations | 匀变速运动方程推导
onstant 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。
ituting 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。
ally, 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。
2. Projectile Motion Trajectory | 抛体运动轨迹方程
lly, 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。
3. Centripetal Acceleration a = v²/r | 向心加速度公式推导
ts 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。
4. Simple Harmonic Motion & Angular Frequency | 简谐运动与角频率
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。
5. Period of a Simple Pendulum | 单摆周期公式推导
gth. 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。
6. Gravitational Field Strength g = GM/r² | 引力场强度推导
eld 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。
7. Electric Field Strength Derivations | 电场强度公式推导
etween 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。
8. Capacitor Discharge Equation | 电容放电方程推导
ives 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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