📚 Mastering Essential Formula Derivations in AS Physics (International) | 国际AS物理核心公式推导精讲
Understanding the derivations behind physics formulas transforms how you approach problem-solving. Rather than relying on rote memorisation, following the logical steps from first principles deepens your conceptual grasp and equips you to tackle unfamiliar questions with confidence. In this article we walk through the essential derivations required for the International AS Physics syllabus, from kinematics to electricity, waves and fields.
理解物理公式的推导过程能够彻底改变你的解题方式。与其机械记忆，不如从基本原理出发，跟随逻辑步骤，这样既能加深概念理解，也能自信地应对陌生问题。本文带你逐一推导国际AS物理大纲中必须掌握的核心公式，涵盖运动学、电学、波动和场等内容。
1. Deriving the Equations of Motion | 运动学方程推导
Consider an object moving with constant acceleration a. The definition of acceleration gives a = (v – u) / t. Rearranging directly yields the first equation.
考虑一个以恒定加速度a运动的物体。根据加速度的定义，有a = (v – u) / t。直接整理即可得到第一个运动方程。
v = u + at
For displacement, the average velocity is (u + v)/2 under constant acceleration. Displacement s equals average velocity multiplied by time.
对于位移，由于加速度恒定，平均速度为 (u + v)/2。位移 s 等于平均速度乘以时间。
s = ½ (u + v)t
Substitute v = u + at into the displacement equation: s = ½ (u + u + at)t = ½ (2u + at)t = ut + ½ at².
将 v = u + at 代入位移方程：s = ½ (u + u + at)t = ½ (2u + at)t = ut + ½ at²。
s = ut + ½ at²
To eliminate t, rearrange the first equation to t = (v – u)/a and insert into s = ½ (u + v)t: s = ½ (u + v)(v – u)/a. Simplifying the product (v – u)(v + u) = v² – u² gives the third equation.
若要消去时间 t，由第一个方程得 t = (v – u)/a，代入 s = ½ (u + v)t：s = ½ (u + v)(v – u)/a。将 (v – u)(v + u) 简化为 v² – u²，便得到第三式。
v² = u² + 2as
2. Impulse-Momentum Theorem | 动量-冲量定理
Newton’s second law is often stated as F = ma. Since acceleration a = Δv / Δt, we can write F = m Δv / Δt.
牛顿第二定律常表达为 F = ma。因为加速度 a = Δv / Δt，于是可写成 F = m Δv / Δt。
Multiply both sides by the time interval Δt: FΔt = m Δv. The left side is impulse, and the right side is the change in momentum Δp = mv – mu.
两边同乘时间 Δt：FΔt = m Δv。等式左边为冲量，右边为动量变化 Δp = mv – mu。
Impulse = FΔt = Δp
This theorem explains that a force applied over a time interval changes an object’s momentum, and it is particularly useful in collision problems.
该定理表明，力在一段时间内的作用会改变物体的动量，在碰撞问题中尤为有用。
3. Work-Energy Principle and Kinetic Energy | 功能原理与动能
Work done by a constant force is W = F · s. For a particle accelerating under a resultant force, the force component along the displacement does work. Assume a constant resultant force F = m a.
恒力做的功为 W = F · s。对于在合力作用下加速的质点，沿位移方向的分力做功。设恒定合力 F = m a。
Using v² = u² + 2as, we can write as = (v² – u²)/2. Substituting into W = m × (v² – u²)/2 gives W = ½mv² – ½mu².
利用 v² = u² + 2as，可得 as = (v² – u²)/2。代入 W = m × (v² – u²)/2，得到 W = ½mv² – ½mu²。
W = ΔEₖ, where Eₖ = ½mv²
The kinetic energy of an object is thus defined as ½mv², and the work done by the resultant force equals the change in kinetic energy.
因此物体的动能定义为 ½mv²，合力做功等于动能的变化量。
4. Gravitational Potential Energy Derivation | 重力势能推导
When an object is lifted slowly at constant speed, the upward force equals its weight mg. The work done against gravity over a vertical height h is W = mgh.
当一个物体以恒定速度缓慢提升时，向上的力等于其重力 mg。克服重力在竖直高度 h 上所做的功为 W = mgh。
This work is stored as gravitational potential energy. Taking the ground as the reference level where Eₚ = 0, the potential energy at height h is:
这些功储存为重力势能。以地面为参考面，设 Eₚ = 0，则高度 h 处的势能为：
Eₚ = mgh
Note that only changes in gravitational potential energy are physically meaningful, and the derivation assumes a uniform gravitational field.
注意，只有重力势能的变化才具有物理意义，且推导假设重力场均匀。
5. Ohm’s Law and Resistivity | 欧姆定律与电阻率
Ohm’s law for a component states that the potential difference V across it is proportional to the current I through it: V = IR, where R is resistance.
一段导体的欧姆定律表明，其两端的电势差 V 与通过的电流 I 成正比：V = IR，R 为电阻。
Resistance depends on the material’s resistivity ρ, length L, and cross-sectional area A. Experiments show R ∝ L and R ∝ 1/A, leading to:
电阻取决于材料的电阻率 ρ、长度 L 和横截面积 A。实验表明 R ∝ L 且 R ∝ 1/A，因此有：
R = ρ L / A
This formula can be rationalised by considering that a longer conductor gives more obstacles to charge flow, while a larger cross-section allows more current for the same voltage.
这一公式可以这样理解：导体越长，电荷流动遇到的阻碍越多；横截面积越大，在相同电压下可通过的电流越大。
6. Electrical Power Formulas | 电功率公式
Power is the rate of energy transfer. In an electrical circuit, the power dissipated in a component is given by the product of the potential difference and current: P = IV.
功率是能量转移的速率。在电路中，元件耗散的功率等于电势差与电流的乘积：P = IV。
Using Ohm’s law V = IR, we can substitute to obtain alternative forms. Replacing V gives P = I(IR) = I²R.
利用欧姆定律 V = IR 进行代换，可得其他形式。用 V 替换得到 P = I(IR) = I²R。
P = I²R
Alternatively, expressing I as V/R yields P = V(V/R) = V²/R.
另一种形式，将 I 表达为 V/R，得到 P = V(V/R) = V²/R。
P = V² / R
These formulas are useful for analysing energy dissipation in resistors and electric devices.
这些公式对分析电阻和电器中的能量耗散十分重要。
7. Deriving the Thin Lens Formula | 薄透镜公式推导
For a thin converging lens, consider an object placed at distance u from the lens and an image formed at distance v. Using similar triangles created by a ray through the centre and a ray parallel to the principal axis, we can set up ratios.
对于薄凸透镜，设物距为 u，像距为 v。利用通过光心的光线和平行于主轴的光线构成的相似三角形，可以建立比例关系。
From a ray diagram, the triangle formed by object height h₀ and distance u is similar to that formed by image height hᵢ and distance v, giving magnification m = hᵢ/h₀ = v/u. Also, using the ray that passes through the focal point, a further similarity leads to hᵢ/h₀ = f/(u – f).
根据光路图，物体高度 h₀ 与物距 u 构成的三角形，相似于像高 hᵢ 与像距 v 构成的三角形，可得放大率 m = hᵢ/h₀ = v/u。同时，利用经过焦点的光线，可得另一相似关系 hᵢ/h₀ = f/(u – f)。
Equating the two expressions for magnification: v/u = f/(u – f). Cross-multiply to get v(u – f) = uf. Expanding and rearranging: uv – vf = uf → uv = uf + vf.
将放大率的两种表达式相等：v/u = f/(u – f)。交叉相乘得 v(u – f) = uf。展开并整理：uv – vf = uf → uv = uf + vf。
Divide through by uvf: 1/f = 1/u + 1/v. This is the Gaussian form of the thin lens equation.
两边同除以 uvf 得到：1/f = 1/u + 1/v。这就是高斯形式的薄透镜方程。
1/f = 1/u + 1/v
8. Wave Speed Equation | 波速公式
A wave transfers energy without transferring matter. The frequency f is the number of complete oscillations per second, and the wavelength λ is the distance advanced by the wave in one complete cycle.
波传递能量而不传递介质。频率 f 是每秒完整振动的次数，波长 λ 是波在一个完整周期内传播的距离。
Therefore, in one second the wave travels a distance equal to f wavelengths. By definition, speed v = distance / time, so in one second:
因此，一秒内波传播的距离等于 f 个波长。根据定义，速度 v = 距离/时间，于一秒内：
v = fλ
This relationship holds for all types of waves, including sound, light, and water waves, provided the medium is uniform.
该关系对所有波均成立，包括声波、光波和水波，前提是介质均匀。
9. Pressure in a Fluid at Rest | 静止流体中的压强
Consider a column of liquid of density ρ, height h, and cross-sectional area A. The weight of the liquid column is mass × g = ρAhg.
考虑一段密度为 ρ、高度为 h、横截面积为 A 的液柱。液柱的重力等于质量乘以 g = ρAhg。
This weight exerts a force on the base of the column. The pressure p due to the liquid alone is the force per unit area: p = ρAhg / A = ρgh.
这个重力作用于液柱底部。液体本身产生的压强 p 为单位面积上的力：p = ρAhg / A = ρgh。
p = ρgh
If an external pressure p₀ is present at the surface, the total pressure at depth h is p = p₀ + ρgh.
若液面存在外部压强 p₀，则深度 h 处的总压强为 p = p₀ + ρgh。
10. Coulomb’s Law and Electric Field Strength | 库仑定律与电场强度
Coulomb’s law quantifies the electrostatic force between two point charges Q₁ and Q₂ separated by distance r: F ∝ Q₁Q₂ / r². Introducing the constant k = 1/(4πε₀) we have:
库仑定律定量描述两电荷间静电力：F ∝ Q₁Q₂ / r²。引入常数 k = 1/(4πε₀)，则有：
F = k Q₁Q₂ / r²
The electric field strength E at a point is defined as the force per unit positive charge placed at that point: E = F/q. Substituting Coulomb’s law for the force on a test charge q due to a source charge Q gives:
电场强度 E 定义为单位正电荷在该点所受的力：E = F/q。将库仑力公式代入试验电荷 q 受力，可得源电荷 Q 产生的电场：
E = k Q / r²
This radial field formula is fundamental for analysing the behaviour of charges and also forms the basis for calculating the uniform field between parallel plates, E = V/d, which follows from work done against the electric force.
这一径向电场公式是分析电荷行为的基础，也是推导平行板间匀强电场 E = V/d 的基础，后者源自电场力做功的关系。
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