📚 AS Physics Unit 1 Formula Derivations – January 2022 Paper Insights | AS物理单元1公式推导——2022年1月真题透视
In the January 2022 AS Physics Unit 1 examination, several questions required candidates to derive key mechanical and material equations from first principles. Mastering these derivations not only secures marks in structured questions but deepens understanding of how physical laws are interconnected. This article revisits the core derivations tested in that paper and explains the logic behind each step.
在2022年1月的AS物理单元1考试中，有多道题目要求考生从基本原理出发推导关键的力学和材料方程。掌握这些推导不仅能确保在结构性题目中得分，还能加深对物理定律之间相互联系的理解。本文回顾了该试卷中考查的核心推导，并解释了每一步背后的逻辑。
1. Defining Acceleration | 加速度的定义
Acceleration is defined as the rate of change of velocity. If an object’s velocity changes from an initial value u to a final value v over a time interval t, the acceleration a is given by the ratio of the change in velocity to the time taken.
加速度定义为速度的变化率。如果一个物体的速度在时间间隔 t 内从初值 u 变为末值 v，则加速度 a 由速度变化量与所用时间的比值给出。
a = (v – u) ÷ t
Rearranging this definition immediately yields the first equation of motion for constant acceleration, which is the foundation for all subsequent derivations.
重新整理这个定义，立刻得到匀加速情况下的第一个运动学方程，这是所有后续推导的基础。
v = u + at
2. Deriving Displacement from Average Velocity | 由平均速度推导位移
For an object moving with uniform acceleration, the velocity changes linearly with time. The average velocity vₐᵥ is therefore the arithmetic mean of the initial and final velocities.
对于做匀加速运动的物体，速度随时间线性变化。因此，平均速度 vₐᵥ 是初速度与末速度的算术平均值。
vₐᵥ = (u + v) ÷ 2
Displacement s is the product of average velocity and time. Substituting the expression for vₐᵥ gives a relation that still contains the final velocity v.
位移 s 是平均速度与时间的乘积。代入平均速度表达式，得到一个仍含有末速度 v 的关系式。
s = (u + v)t ÷ 2
This is a useful intermediate form, especially when time is not directly required. It is often tested in questions that ask for a derivation without eliminating v first.
这是一个有用的中间形式，特别是在不需要直接求解时间的情况下。在要求不先消去 v 而进行推导的题目中经常出现。
3. The Displacement–Time Equation | 位移—时间方程
To express displacement solely in terms of initial velocity, acceleration and time, we substitute v = u + at into s = (u + v)t ÷ 2.
为了只用初速度、加速度和时间表示位移，我们将 v = u + at 代入 s = (u + v)t ÷ 2。
s = (u + (u + at))t ÷ 2 = (2u + at)t ÷ 2
Simplifying the numerator and dividing by 2 yields the standard displacement–time equation for constant acceleration.
简化分子并除以2，得到匀加速运动的标准位移—时间方程。
s = ut + ½at²
This derivation demonstrates the power of algebraic substitution based on the definitions of acceleration and average velocity. It is a staple in both multiple-choice and structured questions.
这一推导展示了基于加速度和平均速度定义的代数代换的威力。它是选择题和结构性题目中的基本内容。
4. The Velocity–Displacement Relation | 速度—位移关系
Sometimes a problem does not provide the time t. By eliminating t from the first two equations of motion, we obtain a direct link between initial velocity, final velocity, acceleration and displacement.
有时问题并未给出时间 t。通过从前两个运动学方程中消去 t，我们可以得到初速度、末速度、加速度和位移之间的直接联系。
From v = u + at, we have t = (v – u) ÷ a. Substituting this into s = (u + v)t ÷ 2 eliminates t and gives an expression that can be rearranged to the familiar form.
由 v = u + at 得 t = (v – u) ÷ a。将其代入 s = (u + v)t ÷ 2 消去 t，得到一个可以整理为熟悉形式的表达式。
s = (u + v)(v – u) ÷ (2a) = (v² – u²) ÷ (2a)
Multiplying both sides by 2a isolates the squared terms, yielding the third equation of motion.
两边同乘 2a，分离平方项，得到第三个运动学方程。
v² = u² + 2as
5. Kinetic Energy from Work Done | 由功推导动能
The kinetic energy formula originates from considering the work done by a resultant force to accelerate a particle from rest. For a constant force F acting over a displacement s, the work done is W = Fs.
动能公式源于考虑合力使粒子从静止加速所做的功。对于一个恒力 F 作用一段位移 s，所做的功为 W = Fs。
Using Newton’s second law, F = ma, and the motion equation v² = 0 + 2as (starting from rest, u = 0), we find s = v² ÷ (2a). Substituting F and s into the work expression eliminates a.
利用牛顿第二定律 F = ma 和运动方程 v² = 0 + 2as（从静止开始，u = 0），求得 s = v² ÷ (2a)。将 F 和 s 代入功的表达式可消去 a。
W = ma × (v² ÷ 2a) = ½mv²
This work is stored as the kinetic energy of the particle, establishing the well-known relationship. The derivation shows how mechanics principles unify force, motion and energy.
这个功以粒子动能的形式储存起来，从而建立了这一众所周知的关系。该推导展示了力学原理如何将力、运动和能量统一起来。
Eₖ = ½mv²
6. Gravitational Potential Energy Near Earth’s Surface | 地表附近的重力势能
When an object of mass m is raised vertically through a height Δh at a constant speed, the lifting force must exactly balance its weight mg. The work done by this lifting force against gravity is Fd = mgΔh.
当质量为 m 的物体以恒定速度垂直升高 Δh 时，举力必须恰好平衡其重力 mg。该举力克服重力所做的功为 Fd = mgΔh。
This work becomes gravitational potential energy stored in the Earth–object system. Hence the change in GPE is simply mgΔh, assuming the gravitational field strength is uniform over the height change.
这个功转化为地球—物体系统中储存的重力势能。因此，假设在高度变化范围内重力场强度均匀，重力势能的变化就是 mgΔh。
ΔEₚ = mgΔh
This derivation is frequently examined in questions that ask for energy conservation in contexts like falling objects or pendulum swings.
在涉及落体或钟摆等情境中的能量守恒问题时，这个推导经常被考查。
7. Instantaneous Power as a Product of Force and Velocity | 瞬时功率是力与速度的乘积
Power is defined as the rate of doing work. For a constant force F causing a small displacement Δs in a time Δt, the average power is P = FΔs ÷ Δt.
功率定义为做功的速率。对于恒力 F 在时间 Δt 内产生微小位移 Δs，平均功率 P = FΔs ÷ Δt。
In the limit as Δt approaches zero, Δs/Δt becomes the instantaneous velocity v. Thus the instantaneous power delivered by a force acting on a moving object is P = Fv, provided F and v are in the same direction.
当 Δt 趋近于零时，Δs/Δt 变为瞬时速度 v。因此，作用在运动物体上的力提供的瞬时功率为 P = Fv，前提是 F 与 v 同方向。
P = Fv
This relationship is useful for analysing constant-power situations, such as a car engine overcoming resistive forces at a steady speed.
该关系对于分析恒定功率情境非常有用，例如汽车发动机在恒定速度下克服阻力的情况。
8. Elastic Potential Energy Stored in a Stretched Spring | 拉伸弹簧中储存的弹性势能
For a spring or a wire obeying Hooke’s law, the force F needed to produce an extension x is directly proportional to x, so F = kx, where k is the spring constant.
对于遵守胡克定律的弹簧或金属丝，产生伸长量 x 所需的力 F 与 x 成正比，即 F = kx，其中 k 为劲度系数。
The work done to stretch the spring is the area under the force–extension graph. Since the graph is a straight line through the origin, the area is a triangle of base x and height kx.
拉伸弹簧所做的功是力—伸长图下的面积。因为图像是一条过原点的直线，该面积是一个底为 x、高为 kx 的三角形。
W = ½ × base × height = ½ × x × kx = ½kx²
This work is stored as elastic potential energy within the spring, provided the elastic limit has not been exceeded.
只要未超过弹性限度，这个功就以弹性势能的形式储存在弹簧中。
Eₑ = ½kx²
9. Young Modulus and its Derivation | 杨氏模量及其推导
The Young modulus E is a measure of the stiffness of a material, defined as the ratio of tensile stress to tensile strain within the Hooke’s law region.
杨氏模量 E 是材料刚度的量度，定义为胡克定律范围内拉伸应力与拉伸应变的比值。
E = tensile stress ÷ tensile strain
Tensile stress is the force F per unit cross-sectional area A, and tensile strain is the extension ΔL divided by the original length L. Substituting these definitions gives the full expression.
拉伸应力是单位横截面积 A 上的力 F，拉伸应变是伸长量 ΔL 除以原始长度 L。代入这些定义可得完整的表达式。
E = (F ÷ A) ÷ (ΔL ÷ L) = (FL) ÷ (AΔL)
This derived formula is particularly useful for calculating the extension of a wire under a known load, and it links the macroscopic deformation to an intrinsic material property.
这一推导出的公式对于计算已知负载下金属丝的伸长量特别有用，并将宏观变形与固有材料属性联系起来。
10. Impulse–Momentum Theorem from Newton’s Second Law | 由牛顿第二定律推导冲量—动量定理
Newton’s second law in its general form states that the resultant force on an object equals the rate of change of its momentum. For constant mass m, this can be written as F = m × a = m × (v – u) ÷ t.
牛顿第二定律的普遍形式指出，物体上的合力等于其动量的变化率。对于恒定质量 m，可写为 F = m × a = m × (v – u) ÷ t。
Multiplying both sides by the time interval t yields the impulse Ft, which equals the change in momentum mv – mu.
两边同乘时间间隔 t，得到冲量 Ft，它等于动量的变化量 mv – mu。
Ft = mv – mu
This relationship, known as the impulse–momentum theorem, explains why airbags and crumple zones increase the impact time to reduce the average force during collisions.
这一关系称为冲量—动量定理，它解释了为什么安全气囊和溃缩区通过延长撞击时间来减小碰撞过程中的平均作用力。
11. Conservation of Momentum from Newton’s Third Law | 由牛顿第三定律推导动量守恒
Consider two bodies A and B that interact for a time Δt. According to Newton’s third law, the force Fᴬᴮ exerted by A on B is equal in magnitude and opposite in direction to the force Fᴮᴬ exerted by B on A.
考虑相互作用的两个物体 A 和 B，作用时间为 Δt。根据牛顿第三定律，A 施加于 B 的力 Fᴬᴮ 与 B 施加于 A 的力 Fᴮᴬ 大小相等、方向相反。
Fᴬᴮ = –Fᴮᴬ
Applying the impulse–momentum theorem to each body, the impulse on A is FᴮᴬΔt = mᴬvᴬ – mᴬuᴬ, and on B is FᴬᴮΔt = mᴮvᴮ – mᴮuᴮ. Adding these two equations shows that the total impulse is zero, so the total momentum is unchanged.
对冲量—动量定理应用于每个物体，A 所受冲量为 FᴮᴬΔt = mᴬvᴬ – mᴬuᴬ，B 所受冲量为 FᴬᴮΔt = mᴮvᴮ – mᴮuᴮ。将两式相加，总冲量为零，因此总动量不变。
mᴬuᴬ + mᴮuᴮ = mᴬvᴬ + mᴮvᴮ
This derivation from Newton’s laws is frequently required in examination questions, especially when asking why momentum is conserved in a closed system.
这种从牛顿定律出发的推导在考试题中经常要求，尤其是当问及为什么在封闭系统中动量守恒时。
12. Summary and Exam Tips | 总结与备考建议
The January 2022 paper highlighted the importance of being able to reproduce these derivations clearly, starting from definitions or fundamental laws. Candidates who simply memorised final equations often lost marks because they could not show the logical flow from first principles to the required relationship.
2022年1月的试卷凸显了能够从定义或基本定律出发清晰复现这些推导的重要性。仅仅记住最终方程式的考生往往因无法展示从基本原理到所需关系式的逻辑流程而丢分。
Practice writing out each derivation step by step, noting the justification for each algebraic manipulation. Pay special attention to the starting assumptions, such as constant acceleration, Hookean behaviour, or constant mass, as these are often the focus of subsequent application questions.
要练习一步步写出每个推导过程，记下每一步代数操作的依据。特别注意起始假设，如匀加速、胡克行为或恒定质量，因为这些常常是后续应用题目的焦点。
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