📚 Formula Derivations from the Edexcel IGCSE Physics Student Book | Edexcel IGCSE 物理学生用书公式推导
Understanding how key formulas are derived is essential for tackling IGCSE Physics problems with confidence. In this article, we walk through step-by-step derivations of the most important equations from the Edexcel IGCSE Physics Student Book. By following these logical progressions, you will strengthen your conceptual grasp and be able to apply each formula correctly under exam conditions.
理解关键公式的推导过程对于自信地解决 IGCSE 物理问题至关重要。在本文中，我们将逐步推导 Edexcel IGCSE 物理学生用书中最重要的方程。通过跟随这些逻辑推导，你将加强概念理解，并能在考试条件下正确应用每个公式。
1. Defining Speed and Acceleration | 速度和加速度的定义
In kinematics, average speed is defined as the total distance travelled divided by the time taken. If an object moves a distance s in time t, its speed v is:
在运动学中，平均速度定义为总运动距离除以所用时间。如果一个物体在时间 t 内移动了距离 s，其速度 v 为：
v = s / t
Acceleration is the rate of change of velocity. When 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:
加速度是速度变化的速率。当物体的速度在时间间隔 t 内从初始值 u 变为最终值 v 时，加速度 a 由下式给出：
a = (v – u) / t
These two definitions form the foundation for all the equations of uniformly accelerated motion that follow.
这两个定义构成了接下来所有匀加速运动方程的基础。
2. First Equation of Motion: v = u + at | 运动第一方程：v = u + at
We start with the definition of acceleration for uniform motion:
我们从匀加速运动的加速度定义出发：
a = (v – u) / t
Multiply both sides by t to obtain:
两边同时乘以 t，得到：
at = v – u
Finally, rearrange by adding u to both sides. This yields the first equation of motion, which expresses final velocity in terms of initial velocity, acceleration and time:
最后，将 u 加到等式两边。这样得出运动第一方程，它用初速度、加速度和时间表示末速度：
v = u + at
3. Second Equation: s = ut + ½at² | 第二方程：s = ut + ½at²
For uniform acceleration, the velocity-time graph is a straight line. The total displacement s is equal to the area under this graph. The area of a trapezium with parallel sides u and v and width t is the average velocity multiplied by time:
对于匀加速运动，速度–时间图是一条直线。总位移 s 等于该图下的面积。以 u 和 v 为平行边、宽度为 t 的梯形面积等于平均速度乘以时间：
s = (u + v) / 2 × t
Now substitute the expression for v from the first equation (v = u + at):
现在将第一方程中的 v 表达式 (v = u + at) 代入：
s = (u + u + at) / 2 × t = (2u + at) / 2 × t
Simplify by distributing t: s = u t + ½ a t². This is the second equation of motion, relating displacement to initial velocity, acceleration and time.
化简并分配 t：s = u t + ½ a t²。这就是运动第二方程，将位移与初速度、加速度和时间联系起来。
4. Third Equation: v² = u² + 2as | 第三方程：v² = u² + 2as
To eliminate time t from the first two equations, start with v = u + at and solve for t:
为了从前两个方程中消去时间 t，由 v = u + at 解出 t：
t = (v – u) / a
Insert this into the second equation s = u t + ½ a t²:
将其代入第二方程 s = u t + ½ a t²：
s = u × (v – u)/a + ½ a × ((v – u)/a)²
Simplify term by term: the first term becomes u(v – u)/a. The second term becomes ½ a × (v – u)²/a² = (v – u)²/(2a). Multiply through by 2a:
逐项化简：第一项变为 u(v – u)/a。第二项变为 ½ a × (v – u)²/a² = (v – u)²/(2a)。两边乘以 2a：
2a s = 2u(v – u) + (v – u)²
Expand and collect terms: 2a s = 2uv – 2u² + v² – 2uv + u² = v² – u². Rearranging gives the third equation of motion:
展开并合并同类项：2a s = 2uv – 2u² + v² – 2uv + u² = v² – u²。移项得到运动第三方程：
v² = u² + 2as
5. Newton’s Second Law and F = ma | 牛顿第二定律与 F = ma
Newton’s second law states that the resultant force on an object is directly proportional to the rate of change of its momentum. Momentum p is defined as mass × velocity: p = m v. For an object with constant mass m, the change in momentum over a short time interval Δt is Δp = m Δv. Therefore:
牛顿第二定律指出，物体所受的合力与其动量的变化率成正比。动量 p 定义为质量 × 速度：p = m v。对于质量 m 恒定的物体，在短时间 Δt 内动量的变化为 Δp = m Δv。因此：
F ∝ Δp / Δt = m Δv / Δt = m a
Using SI units, the proportionality constant is set to 1, giving the familiar vector equation:
使用国际单位制，比例常数被设为 1，从而得到熟悉的矢量方程：
F = m a
This equation relates net force, mass and acceleration and is a cornerstone of dynamics.
这个方程将合力、质量和加速度联系起来，是动力学的基石。
6. Work and Kinetic Energy | 功和动能
When a constant net force F moves an object through a displacement s in the direction of the force, the work done W is W = F s. For an object of mass m accelerating uniformly from rest (u = 0), we can substitute F = m a and use the third equation of motion with u = 0: v² = 2 a s → s = v² / (2a).
当恒定的合力 F 使物体沿力的方向发生位移 s 时，所做的功 W 为 W = F s。对于由静止 (u = 0) 开始匀加速的质量为 m 的物体，我们可以代入 F = m a，并使用初速度为零的运动第三方程：v² = 2 a s → s = v² / (2a)。
W = m a × (v² / (2a)) = ½ m v²
This work is entirely converted into kinetic energy. Hence the kinetic energy Eₖ of a moving object is:
这些功全部转化为动能。因此，运动物体的动能 Eₖ 为：
Eₖ = ½ m v²
7. Gravitational Potential Energy | 重力势能
To lift an object of mass m vertically through a height h near the Earth’s surface, you must do work against the gravitational force m g, where g is the acceleration of free fall. The minimum force required equals the weight, and the work done in lifting it is:
要在地球表面附近将质量为 m 的物体竖直升高 h，你必须克服重力 m g 做功，其中 g 是自由落体加速度。所需的最小力等于重力，将其举高所做的功为：
W = F d = (m g) × h = m g h
Assuming no other energy transfers, this work is stored as gravitational potential energy. Therefore, the change in gravitational potential energy ΔEₚ is:
假设没有其他能量转换，这些功被储存为重力势能。因此，重力势能的变化量 ΔEₚ 为：
ΔEₚ = m g h
This expression gives the energy stored relative to the initial height.
这个表达式给出了相对于初始高度储存的能量。
8. Pressure in a Liquid | 液体压强
Consider a column of liquid of density ρ, height h and cross-sectional area A. Its volume is V = A h, so its mass is m = ρ V = ρ A h. The weight of the column is W = m g = ρ A h g.
考虑一个密度为 ρ、高度为 h、横截面积为 A 的液柱。其体积为 V = A h，因此质量为 m = ρ V = ρ A h。液柱的重量为 W = m g = ρ A h g。
Pressure p is defined as force per unit area. The force exerted on the base is the weight of the column, so:
压强 p 定义为单位面积上的力。作用在底部的力就是液柱的重量，因此：
p = W / A = (ρ A h g) / A = ρ g h
This shows that the pressure at a depth h in a static liquid depends only on the liquid’s density, the acceleration due to gravity and the depth, not on the cross-sectional area.
这表明，在静止液体中深度 h 处的压强仅取决于液体的密度、重力加速度和深度，而与横截面积无关。
9. Resistance in Series and Parallel | 串联与并联电阻
Series circuit: The same current I flows through each resistor. The total potential difference V across the series combination is the sum of the individual p.d.s: V = V₁ + V₂. Using Ohm’s law, V₁ = I R₁ and V₂ = I R₂, so V = I R₁ + I R₂ = I (R₁ + R₂). For the equivalent resistance Rₛ, V = I Rₛ. Hence:
串联电路：相同的电流 I 流过每个电阻。串联组合两端的总电势差 V 等于各电势差之和：V = V₁ + V₂。利用欧姆定律，V₁ = I R₁ 且 V₂ = I R₂，因此 V = I R₁ + I R₂ = I (R₁ + R₂)。对于等效电阻 Rₛ，V = I Rₛ。因此：
Rₛ = R₁ + R₂
Parallel circuit: The same potential difference V appears across each branch. The total current I from the source splits: I = I₁ + I₂. Applying Ohm’s law to each branch gives I₁ = V / R₁ and I₂ = V / R₂. Thus I = V / R₁ + V / R₂ = V (1/R₁ + 1/R₂). For the equivalent parallel resistance Rₚ, I = V / Rₚ. Equating:
并联电路：每个支路两端的电势差 V 相同。来自电源的总电流 I 分流：I = I₁ + I₂。对每个支路应用欧姆定律得 I₁ = V / R₁，I₂ = V / R₂。因此 I = V / R₁ + V / R₂ = V (1/R₁ + 1/R₂)。对于等效并联电阻 Rₚ，I = V / Rₚ。两式相等：
1/Rₚ = 1/R₁ + 1/R₂
These relationships can be extended to more than two resistors.
这些关系可以扩展到两个以上的电阻。
10. Electrical Power Equations | 电功率公式
Published by TutorHao | IGCSE Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导