📚 Deriving Key Formulas from Cambridge IGCSE® O Level Complete Physics | 剑桥IGCSE® O Level完整物理公式推导
The Cambridge IGCSE® O Level Complete Physics Student Book Fourth Edition presents physics not as a collection of isolated facts but as a coherent subject built from fundamental principles. Understanding how key formulas are derived strengthens conceptual clarity, reduces rote memorisation, and equips learners to tackle unfamiliar problems with confidence. In this article, we walk through the derivations of the essential equations that form the backbone of the syllabus.
剑桥 IGCSE® O Level 完整物理学生用书第四版将物理呈现为一个由基本原理构建而成的连贯学科，而非孤立的知识点集合。理解核心公式的推导过程能强化概念清晰度，减少死记硬背，并使学习者能够自信地应对陌生问题。本文将逐一梳理构成课程大纲支柱的关键方程式的推导。
1. Defining Speed and Acceleration | 定义速度与加速度
Average speed is defined as the total distance travelled divided by the total time taken. If an object moves a distance d in a time t, we write:
平均速度定义为总移动距离除以总时间。若物体在时间 t 内移动了距离 d，可写作：
average speed = d / t
This definition can be rearranged to find distance, d = average speed × t, which is useful for analysing uniform motion. For motion with changing velocity, we introduce acceleration. Acceleration a is the rate of change of velocity. If the velocity increases from an initial value u to a final value v over a time t, the acceleration is given by:
此定义可变形以求出距离 d = 平均速度 × t，这对于分析匀速运动十分有用。对于速度变化的运动，我们引入加速度。加速度 a 是速度的变化率。若速度在时间 t 内从初值 u 增加到末值 v，加速度为：
a = (v – u) / t
This formula assumes constant acceleration. Rearranging gives v = u + a t, which is our first equation of motion and a direct consequence of the definition of acceleration.
此公式假设加速度恒定。移项后得到 v = u + a t，这是我们的第一个运动方程，也是加速度定义的直接结果。
2. Deriving the Equations of Motion | 运动方程的推导
When acceleration is constant, a velocity–time graph is a straight line. The area under the graph represents displacement s. The area is a trapezium, so we can calculate s as the product of the average velocity and time:
当加速度恒定时，速度-时间图像是一条直线。图像下的面积代表位移 s。该区域为梯形，因此可用平均速度与时间的乘积计算 s：
s = ½(u + v) t
Substituting v = u + a t into this displacement formula yields s = ½(u + u + a t) t, which simplifies to:
将 v = u + a t 代入此位移公式得到 s = ½(u + u + a t) t，化简得：
s = u t + ½ a t²
If we eliminate t from s = ½(u + v)t using t = (v – u)/a, we obtain s = ½(u + v)(v – u)/a = (v² – u²) / (2a). Rearranging gives the fourth useful relationship:
若用 t = (v – u)/a 从 s = ½(u + v)t 中消去 t，可得 s = ½(u + v)(v – u)/a = (v² – u²) / (2a)。移项后得到第四个有用的关系式：
v² = u² + 2 a s
These four equations of motion – v = u + a t, s = ½(u + v) t, s = u t + ½ a t², and v² = u² + 2 a s – are used extensively to solve problems involving constant acceleration in a straight line.
这四个运动方程——v = u + a t、s = ½(u + v) t、s = u t + ½ a t² 以及 v² = u² + 2 a s——广泛用于解决匀加速直线运动的问题。
3. Newton’s Second Law and F = ma | 牛顿第二定律与 F = ma
Newton’s second law states that the net force acting on an object is directly proportional to the rate of change of its momentum. When mass is constant, this is written as F ∝ m a. Through experimentation, we find that doubling the force doubles the acceleration for a fixed mass, and doubling the mass halves the acceleration for a fixed force. By defining the unit of force, the newton (N), as the force that gives a mass of 1 kg an acceleration of 1 m/s², the proportionality constant becomes 1:
牛顿第二定律指出，作用于物体的净力与其动量的变化率成正比。当质量恒定时，可写为 F ∝ m a。实验表明，质量一定时力加倍则加速度加倍，力一定时质量加倍则加速度减半。通过定义力的单位牛顿 (N) 为使 1 kg 物体产生 1 m/s² 加速度的力，比例常数变为 1：
F = m a
This simple relationship connects force, mass and acceleration. It is the foundation for dynamics. For example, the weight of an object is the force due to gravity on its mass: W = m g, where g is the acceleration of free fall near the Earth’s surface, approximately 9.8 m/s².
这一简洁的关系式将力、质量和加速度联系起来，是动力学的基础。例如，物体的重量就是重力作用于其质量的力：W = m g，其中 g 为地球表面附近的自由落体加速度，约为 9.8 m/s²。
4. Momentum and Impulse | 动量与冲量
Momentum p is defined as the product of mass and velocity: p = m v. From Newton’s second law in its more general form, force equals the rate of change of momentum: F = Δp / Δt. For constant mass, this becomes F = (m v – m u) / Δt = m (v – u)/Δt = m a, consistent with F = m a. Rearranging gives the impulse–momentum relationship:
动量 p 定义为质量与速度的乘积：p = m v。根据牛顿第二定律的更一般形式，力等于动量的变化率：F = Δp / Δt。当质量不变时，变为 F = (m v – m u) / Δt = m (v – u)/Δt = m a，与 F = m a 一致。移项后得到冲量-动量的关系：
F t = m v – m u = Δ p
The product of force and time, F t, is called impulse. This equation tells us that the impulse acting on an object equals the change in its momentum. It is especially useful when forces act over very short time intervals, such as in collisions.
力与时间的乘积 F t 称为冲量。该方程表明，作用于物体的冲量等于其动量的变化量。在碰撞等力作用时间极短的情况下，这一关系尤为有用。
5. Work, Kinetic Energy and Potential Energy | 功、动能与势能
Work is done when a force moves an object. The work done W by a constant force F moving a distance d in the direction of the force is:
当力使物体移动时，力就做了功。恒力 F 在力的方向上移动距离 d 所做的功为：
W = F d
Consider a constant net force accelerating an object from rest. Using F = m a and v² = u² + 2 a s with u = 0, we get a s = v² / 2. Then:
考虑一个恒定的净力使物体从静止开始加速。利用 F = m a 和 v² = u² + 2 a s，且 u = 0，得到 a s = v² / 2。于是：
W = F s = m a s = m (v² / 2) = ½ m v²
This work done on the object is stored as kinetic energy (KE). Hence the kinetic energy of an object of mass m moving at speed v is KE = ½ m v². If the object already had an initial speed u, the work done equals the change in kinetic energy: F d = ½ m v² – ½ m u².
对物体做的这个功被储存为动能 (KE)。因此，质量为 m、速度为 v 的物体具有动能 KE = ½ m v²。若物体已有初速度 u，则所做的功等于动能的变化量：F d = ½ m v² – ½ m u²。
When lifting an object through a vertical height h, the force required to overcome gravity is the weight m g. The work done against gravity is W = m g h. This energy becomes gravitational potential energy (GPE):
当克服重力将物体竖直提升高度 h 时，所需的力为重量 m g。克服重力所做的功为 W = m g h。这部分能量转化为重力势能 (GPE)：
GPE = m g h
In the absence of air resistance and friction, the decrease in GPE equals the increase in KE, leading to the principle of conservation of mechanical energy.
在无空气阻力和摩擦的情况下，重力势能的减少量等于动能的增加量，从而得出机械能守恒原理。
6. Pressure in Fluids: p = ρgh | 流体压强：p = ρgh
Pressure p is defined as the force acting perpendicularly per unit area: p = F / A. In a liquid, the pressure at a depth h arises from the weight of the liquid column above that point. Consider a column of liquid of density ρ, cross-sectional area A, and height h. Its volume is V = A h, its mass is m = ρ V = ρ A h, and its weight is W = m g = ρ A h g.
压强 p 定义为垂直作用于单位面积上的力：p = F / A。在液体中，深度 h 处的压强源自该点上方液柱的重量。考虑一个密度为 ρ、截面积为 A、高度为 h 的液柱。其体积 V = A h，质量 m = ρ V = ρ A h，重量 W = m g = ρ A h g。
Since the force acting on the area A at depth h is this weight, the pressure due to the liquid alone is:
由于作用在深度 h 处面积 A 上的力就是该重量，因此仅由液体产生的压强为：
p = F / A = (ρ A h g) / A = ρ g h
This is the hydrostatic pressure equation. The total pressure at a point in a liquid includes the atmospheric pressure acting on the surface, so total p = p_atm + ρ g h. The derivation shows that liquid pressure depends only on depth, density, and gravitational field strength, not on the total volume or shape of the container.
这就是流体静压强的方程。液体中某点的总压强包含作用在液面上的大气压强，因此总压强 p = p_atm + ρ g h。该推导表明液体压强仅取决于深度、密度和重力场强度，而与液体的总体积或容器形状无关。
7. Specific Heat Capacity and Heat Transfer | 比热容与热传递
When energy is transferred to a substance, its temperature may rise. The specific heat capacity c is the energy required to raise the temperature of 1 kg of the substance by 1 °C (or 1 K). For a mass m experiencing a temperature change Δθ, the heat transferred Q is:
当能量传递给物质时，其温度可能升高。比热容 c 是使 1 kg 物质温度升高 1 °C（或 1 K）所需的能量。对于发生温度变化 Δθ 的质量 m，传递的热量 Q 为：
Q = m c Δθ
This formula is derived from the definition of specific heat capacity and the principle of conservation of energy: electrical energy supplied (V I t) is equated to the thermal energy gained (m c Δθ), assuming no heat losses. If a heater of power P delivers energy for a time t, then P t = m c Δθ, allowing c to be determined experimentally.
该公式由比热容的定义和能量守恒原理推导得出：假设无热量损失，供给的电能 (V I t) 等于获得的热能 (m c Δθ)。若一个功率为 P 的加热器提供能量 t 秒，则有 P t = m c Δθ，从而可通过实验测定 c。
During a change of state, the temperature remains constant, and the energy supplied goes into breaking inter-particle bonds rather than increasing kinetic energy. The specific latent heat L is the energy required to change the state of 1 kg of a substance without a temperature change:
在物态变化过程中，温度保持不变，所提供的能量用于打破粒子间的键，而非增加动能。比潜热 L 是使 1 kg 物质在无温度变化时改变物态所需的能量：
Q = m L
Both equations are fundamental in calorimetry and for understanding energy budgets in heating and cooling processes.
这两个方程在量热学中以及在理解加热和冷却过程中的能量平衡方面都是基础。
8. Ohm’s Law and Resistance in Circuits | 欧姆定律与电路中的电阻
Ohm’s law states that the current I through a metallic conductor is directly proportional to the potential difference V across it, provided temperature remains constant. The constant of proportionality is the resistance R:
欧姆定律指出，在温度恒定的条件下，通过金属导体的电流 I 与其两端的电势差 V 成正比。比例常数即为电阻 R：
V = I R
This defines resistance as R = V / I. For resistors connected in series, the current is the same through each. The total potential difference is V_total = V₁ + V₂ + … . Using V = I R for each, we have I R_total = I R₁ + I R₂ + … . Cancelling the common current I gives:
这一定义了电阻 R = V / I。对于串联的电阻器，流过每个电阻器的电流相同。总电势差 V_total = V₁ + V₂ + … 。对每个电阻器使用 V = I R，得到 I R_total = I R₁ + I R₂ + … 。约去公共电流 I，得：
R_total = R₁ + R₂ + …
For resistors in parallel, the potential difference across each branch is the same. The total current from the source splits: I_total = I₁ + I₂ + … . Substituting I = V / R gives V / R_total = V / R₁ + V / R₂ + … . Cancelling V yields:
对于并联的电阻器，每个支路两端的电势差相同。从电源流出的总电流分流：I_total = I₁ + I₂ + … 。代入 I = V / R，得到 V / R_total = V / R₁ + V / R₂ + … 。约去 V，得：
1 / R_total = 1 / R₁ + 1 / R₂ + …
These derivations rely only on Ohm’s law, the conservation of charge (current) at junctions, and the conservation of energy (potential difference) around a closed loop, which are the foundations of circuit analysis.
这些推导仅依赖于欧姆定律、节点处的电荷守恒（电流）以及闭合回路中的能量守恒（电势差），这也是电路分析的基础。
9. Electrical Power and Energy Dissipation | 电功率与能量消耗
Electrical power P is the rate at which electrical energy is transferred. By definition, power is work done per unit time. When a charge Q moves through a potential difference V, the work done is W = V Q. Current is I = Q / t, so Q = I t. Substituting gives W = V I t. Therefore, power is:
电功率 P 是电能转换的速率。根据定义，功率是单位时间内所做的功。当电荷 Q 通过电势差 V 时，做功为 W = V Q。电流 I = Q / t，因此 Q = I t。代入得到 W = V I t。于是功率为：
P = W / t = V I
For a component that obeys Ohm’s law, we can substitute V = I R or I = V / R to obtain alternative forms:
对于服从欧姆定律的元件，可代入 V = I R 或 I = V / R，得到其他形式：
P = I² R and P = V² / R
The energy transferred in a time t is E = P t = V I t = I² R t = (V² / R) t. These equations are used to calculate the heating effect in resistors, the brightness of lamps, and the energy consumption of household appliances, often expressed in kilowatt-hours (kWh).
在时间 t 内传递的能量为 E = P t = V I t = I² R t = (V² / R) t。这些等式用于计算电阻器的热效应、灯泡的亮度以及家用电器的能量消耗，能量通常以千瓦时 (kWh) 表示。
10. The Transformer Equation | 变压器公式
A transformer consists of two coils wound on a common iron core. An alternating current in the primary coil produces a changing magnetic flux, which is almost entirely linked through the secondary coil. According to Faraday’s law, the induced e.m.f. across each coil is proportional to the rate of change of flux and the number of turns. Since the same rate of change of flux is experienced by each turn, the e.m.f. per turn is constant. Hence:
变压器由绕在同一铁芯上的两个线圈组成。初级线圈中的交变电流产生变化的磁通量，该磁通量几乎完全通过次级线圈。根据法拉第定律，每个线圈中的感应电动势与磁通量变化率及匝数成正比。由于每匝经历的磁通量变化率相同，因此每匝的电势为常数。从而：
V_p / N_p = V_s / N_s ⇒ V_p / V_s = N_p / N_s
where V_p and V_s are the primary and secondary voltages, and N_p and N_s are the respective numbers of turns. For an ideal transformer with 100% efficiency, the input power equals the output power:
其中 V_p 和 V_s 分别为初级和次级电压，N_p 和 N_s 为各自匝数。对于效率为 100% 的理想变压器，输入功率等于输出功率：
P_in = P_out ⇒ V_p I_p = V_s I_s
Combining the two relations gives I_s / I_p = N_p / N_s, showing that a step-up transformer increases voltage but decreases current proportionally, and vice versa. This derivation is central to understanding power transmission in the national grid.
结合这两个关系式可得 I_s / I_p = N_p / N_s，表明升压变压器升高电压但按比例减小电流，反之亦然。这一推导对于理解国家电网中的电力传输至关重要。
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