📚 AQA International A Level Physics: Key Formula Derivations | AQA国际A Level物理核心公式推导
In AQA International A Level Physics, understanding how key formulas are derived is essential for mastering concepts and excelling in examination questions that require justification, proof, or application from first principles. This article presents a curated selection of core derivations spanning mechanics, waves, electricity, and nuclear physics, laid out step by step with parallel English‑Chinese explanations.
在AQA国际A Level物理中，理解关键公式的推导过程对于掌握概念和在需要论证、证明或从基本原理出发应用的考题中取得优异成绩至关重要。本文精选了横跨力学、波动、电学和核物理的核心推导，以逐步英中对照说明的方式呈现。
1. Derivation of SUVAT Equations | 匀加速运动方程推导
For an object moving with uniform acceleration a, the acceleration is defined as the rate of change of velocity:
对于匀加速运动的物体，加速度定义为速度的变化率：
a = (v − u) / t
Rearranging this definition gives the first SUVAT equation, relating final velocity v, initial velocity u, acceleration a and time t.
整理这一定义得到第一个匀加速运动方程，联系末速度 v、初速度 u、加速度 a 及时间 t。
v = u + a t
The displacement s during the time interval can be found from the area under a velocity‑time graph. For constant acceleration the area is a trapezium, giving the average velocity as ½ (u + v):
该时间间隔内的位移 s 可由速度‑时间图下方的面积求得。对于匀加速度，该面积为梯形，因此平均速度为 ½ (u + v)：
s = ½ (u + v) t
Substituting v = u + a t into the displacement equation yields the form commonly used when the final velocity is unknown.
将 v = u + a t 代入位移方程，得到末速度未知时常用的形式。
s = u t + ½ a t²
Finally, eliminating t from v = u + a t and s = ½ (u + v) t gives the relation linking velocities, acceleration and displacement without explicit time dependence.
最后，从 v = u + a t 和 s = ½ (u + v) t 中消去 t，得到不显含时间的速度、加速度与位移关系式。
v² = u² + 2 a s
2. Derivation of Kinetic Energy Formula | 动能公式推导
Starting from the work–energy principle, the work done by a constant net force F over a displacement s is W = F s. Using Newton’s second law F = m a and the SUVAT equation v² = u² + 2 a s, we can express displacement in terms of the velocity change:
从功能原理出发，恒定合外力 F 在位移 s 上所做的功为 W = F s。利用牛顿第二定律 F = m a 和匀加速方程 v² = u² + 2 a s，可将位移用速度变化表示：
s = (v² − u²) / (2 a)
Substituting into the work expression:
代入功的表达式：
W = m a × (v² − u²) / (2 a) = ½ m v² − ½ m u²
The quantity ½ m v² is defined as the kinetic energy Eₖ. Thus, the work done by the net force equals the change in kinetic energy.
定义 ½ m v² 为动能 Eₖ。因此，合外力所做的功等于动能的变化量。
3. Momentum and Impulse Relationship | 动量与冲量关系推导
Newton’s second law can be written in terms of momentum p = m v. If the mass is constant, the rate of change of momentum is:
牛顿第二定律可以用动量 p = m v 来表示。若质量恒定，动量的变化率为：
F = d p / d t = m (d v / d t) = m a
Multiplying both sides by a small time interval Δ t gives the impulse F Δ t:
两边同乘以微小时间间隔 Δ t，得到冲量 F Δ t：
F Δ t = Δ p = m v − m u
This derivation shows that impulse is equal to the change in momentum, which directly leads to the principle of conservation of momentum when no external force acts.
这一推导表明冲量等于动量的变化量，从而直接得出在没有外力作用时的动量守恒原理。
4. Derivation of Centripetal Acceleration | 向心加速度推导
Consider an object moving at constant speed v in a circle of radius r. In a short time Δ t, the object moves through a small angle Δ θ. The velocity vector changes direction by Δ θ, while its magnitude remains v. The change in velocity Δ v points toward the centre and for small angles has magnitude Δ v ≈ v Δ θ.
考虑一物体以恒定速率 v 在半径为 r 的圆周上运动。在短时间 Δ t 内，物体转过小角度 Δ θ。速度矢量方向改变 Δ θ，大小保持为 v。速度变化量 Δ v 指向圆心，小角度下其大小为 Δ v ≈ v Δ θ。
The acceleration magnitude is therefore:
因此加速度大小为：
a = Δ v / Δ t = v (Δ θ / Δ t) = v ω
Using the relationship between angular speed and linear speed, v = ω r, we obtain two equivalent expressions for centripetal acceleration.
利用角速度与线速度的关系 v = ω r，我们得到向心加速度的两个等价表达式。
a = v² / r = ω² r
5. Simple Harmonic Motion: Displacement Equation | 简谐运动位移方程推导
An object undergoes simple harmonic motion when the restoring force is proportional to the displacement from equilibrium and directed opposite to it: F = − k x. Applying Newton’s second law:
当回复力与偏离平衡位置的位移成正比且方向相反时，物体做简谐运动：F = − k x。应用牛顿第二定律：
m a = − k x → a = − (k / m) x
Defining the angular frequency ω = √(k / m), the acceleration can be written as a = − ω² x. Since acceleration is the second derivative of displacement, this gives the defining differential equation:
定义角频率 ω = √(k / m)，加速度可写为 a = − ω² x。因为加速度是位移的二阶导数，得到定义微分方程：
d² x / d t² = − ω² x
A general solution to this equation is x = A cos (ω t + φ), where A is the amplitude and φ the phase constant. Differentiating gives the velocity and acceleration functions that confirm the motion is sinusoidal.
该方程的一个通解为 x = A cos (ω t + φ)，其中 A 为振幅，φ 为初相。对其求导可得到速度和加速度函数，验证运动是正弦式的。
6. Derivation of Period of a Simple Pendulum | 单摆周期推导
For a simple pendulum of length L displaced by a small angle θ, the restoring force along the arc is the tangential component of weight: F = − m g sin θ. For small angles, sin θ ≈ θ ≈ x / L, where x is the arc length displacement.
对于长度为 L 的单摆，偏离小角度 θ 时，沿弧线的回复力为重力的切向分量：F = − m g sin θ。对于小角度，sin θ ≈ θ ≈ x / L，其中 x 为弧长位移。
Thus, F ≈ − (m g / L) x. This has the same form as the simple harmonic restoring force F = − k x, with effective spring constant k = m g / L. The angular frequency is:
因此，F ≈ − (m g / L) x。这与简谐回复力 F = − k x 形式相同，等效劲度系数为 k = m g / L。角频率为：
ω = √(k / m) = √(g / L)
The period T = 2 π / ω follows directly.
周期 T = 2 π / ω 直接得出。
T = 2 π √(L / g)
7. Capacitor Discharge Equation | 电容器放电方程推导
Consider a capacitor of capacitance C discharging through a resistor R. At any instant, the potential difference across the capacitor is V = Q / C, and the same voltage appears across the resistor: V = I R. The current is the rate at which charge leaves the capacitor, so I = − d Q / d t.
考虑电容为 C 的电容器通过电阻 R 放电。在任意时刻，电容器两端电势差为 V = Q / C，且电阻两端电压与之相同：V = I R。电流是电荷离开电容器的速率，因此 I = − d Q / d t。
Combining these gives the differential equation:
联立上述关系得到微分方程：
− d Q / d t = Q / (R C)
Separating variables and integrating:
分离变量并积分：
∫ d Q / Q = − ∫ d t / (R C) → ln Q = − t / (R C) + constant
Applying the initial condition Q = Q₀ at t = 0 yields the exponential decay law.
应用初始条件 t = 0 时 Q = Q₀，得到指数衰减规律。
Q = Q₀ e⁻ᵗ / (R C)
Corresponding equations for voltage and current follow by substituting V = Q / C and I = V / R.
相应的电压和电流方程可通过代入 V = Q / C 和 I = V / R 得到。
8. Derivation of Half-Life in Radioactive Decay | 放射性衰变半衰期推导
Radioactive decay is a random process where the number of nuclei N decreases at a rate proportional to the number present:
放射性衰变是一个
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