📚 Formula Derivation in AS Physics Paper 1 (January 2018) | AS物理试卷1公式推导（2018年1月）
In AS Physics Paper 1, especially in the January 2018 sitting, candidates often encounter questions that require a solid understanding of how fundamental equations are derived. Being able to reconstruct these derivations not only strengthens conceptual grasp but also helps tackle multiple-choice and structured questions with confidence. This article revisits the key derivations across mechanics, materials, waves and electricity that are central to the AS syllabus.
在AS物理试卷1中，尤其是2018年1月的考试，常会出现需要扎实理解基本方程如何推导的题目。能够自己推导这些公式不仅能加深概念理解，还能自信地应对选择题与结构化问题。本文重温了力学、材料、波和电学中核心公式的推导，这些都是AS阶段的学习重点。
1. Deriving the First Equation of Motion (v = u + at) | 推导第一个运动学方程 (v = u + at)
v − u) / t, where u is the initial velocity, v is the final velocity and t is the time taken.
加速度定义为速度的变化率。对于匀加速运动，这可以写成 a = (v − u) / t，其中 u 是初速度，v 是末速度，t 是所用时间。
h sides by t gives at = v − u. Adding u to both sides yields the first equation of motion.
两边同乘以 t 得到 at = v − u。再在两边加上 u，就得到了第一个运动学方程。
v = u + at
This linear relationship shows that final velocity is simply initial velocity plus the velocity gained from acceleration over time.
这个线性关系表明，末速度等于初速度加上加速度在时间内增加的速度。
2. Deriving the Second Equation of Motion (s = ut + ½at²) | 推导第二个运动学方程 (s = ut + ½at²)
average velocity is (u + v)/2. From the first equation, v = u + at, so substituting gives:
对于匀加速运动，位移 s 等于平均速度乘以时间。平均速度为 (u + v)/2。由第一个方程 v = u + at 代入得到：
Average velocity = (u + u + at)/2 = u + ½at. Multiplying by t gives the displacement.
平均速度 = (u + u + at)/2 = u + ½at。乘以时间 t 即得位移。
s = ut + ½at²
This derivation shows how displacement depends on initial velocity, time and acceleration. The term ½at² arises from the additional distance covered due to acceleration.
这个推导表明位移依赖于初速度、时间和加速度。项 ½at² 来自于因加速而多走的距离。
3. Deriving the Third Equation of Motion (v² = u² + 2as) | 推导第三个运动学方程 (v² = u² + 2as)
+ at)² = u² + 2uat + a²t². Factor out 2a from the last two terms: v² = u² + 2a(ut + ½at²).
先将第一个方程平方：v² = (u + at)² = u² + 2uat + a²t²。将后两项提取公因子 2a：v² = u² + 2a(ut + ½at²)。
Notice that the bracket is precisely the expression for s. Hence, v² = u² + 2as.
注意到括号内正好是 s 的表达式。因此得到 v² = u² + 2as。
v² = u² + 2as
This equation is particularly useful when time is not involved, linking velocity, acceleration and displacement directly.
这个方程在不需要时间时尤其有用，直接关联了速度、加速度和位移。
4. Deriving Kinetic Energy (Eₖ = ½mv²) | 推导动能公式 (Eₖ = ½mv²)
nstant force F. The work done by the force is W = F × s. From Newton’s second law, F = ma.
考虑质量为 m 的物体，在恒力 F 作用下由静止加速到速度 v，位移为 s。力做的功 W = F × s。由牛顿第二定律，F = ma。
² = u² + 2as with u=0 gives v² = 2as, so s = v²/(2a). Substituting into the work equation:
利用 v² = u² + 2as，且 u=0，得 v² = 2as，所以 s = v²/(2a)。代入功的公式：
W = ma × (v²/(2a)) = ½mv². This work is stored as kinetic energy.
W = ma × (v²/(2a)) = ½mv²。这个功以动能的形式储存。
Eₖ = ½mv²
5. Deriving Gravitational Potential Energy (ΔEₚ = mgΔh) | 推导重力势能变化 (ΔEₚ = mgΔh)
force must equal the weight mg. The work done by this force is force × distance = mg × Δh.
将质量为 m 的物体匀速提升竖直高度 Δh，向上的力必须等于重力 mg。该力做的功为力 × 距离 = mg × Δh。
Work done against gravity is stored as gravitational potential energy, so the change in GPE is:
克服重力做的功储存为重力势能，因此重力势能的变化为：
ΔEₚ = mgΔh
This assumes a uniform gravitational field and no other energy transfers.
这里假设均匀重力场且无其他能量转化。
6. Deriving Conservation of Momentum from Newton’s Third Law | 从牛顿第三定律推导动量守恒
Consider two bodies A and B colliding. During collision, A exerts force F on B for time Δt. By Newton’s third law, B exerts force −F on A for the same duration.
考虑两个物体 A 和 B 碰撞。碰撞期间，A 对 B 施加力 F 持续 Δt 时间。根据牛顿第三定律，B 施加力 −F 于 A，时间相同。
Impulse on B = FΔt = mBvB − mBuB. Impulse on A = −FΔt = mAvA − mAuA. Adding these gives:
对 B 的冲量 = FΔt = mBvB − mBuB。对 A 的冲量 = −FΔt = mAvA − mAuA。两式相加：
0 = mAvA + mBvB − (mAuA + mBuB). Hence total momentum before equals total momentum after:
0 = mAvA + mBvB − (mAuA + mBuB)。因此碰撞前总动量等于碰撞后总动量：
mAuA + mBuB = mAvA + mBvB
This derivation confirms that momentum is conserved in an isolated system where no external forces act.
这一推导确认，在无外力作用的孤立系统中，动量守恒。
7. Deriving Elastic Potential Energy (E = ½FΔx or ½k(Δx)²) | 推导弹性势能 (E = ½FΔx 或 ½k(Δx)²)
e k is the spring constant. The force is not constant; it increases linearly from 0 to kx.
对于遵循胡克定律的弹簧，力 F 与伸长量 x 成正比：F = kx，其中 k 是弹簧常数。这个力不是恒定的；它从 0 线性增加到 kx。
The work done in stretching the spring is the area under the force– × base × height = ½ × x × kx = ½kx². Using F = kx, the same energy can be written as ½FΔx.
拉伸弹簧做的功等于力–伸长量图下的面积，这是一个三角形：W = ½ × 底 × 高 = ½ × x × kx = ½kx²。利用 F = kx，同样的能量可写成 ½FΔx。
E = ½FΔx = ½k(Δx)²
This stored energy is recoverable as long as the elastic limit is not exceeded.
只要不超过弹性极限，这部分储存的能量是可恢复的。
8. Deriving Total Resistance for Series Resistors (R = R₁ + R₂) | 推导串联电阻的总电阻 (R = R₁ + R₂)
is the sum of individual p.d.s: V = V₁ + V₂. Using Ohm’s law, V = IR, V₁ = IR₁, V₂ = IR₂.
在串联电路中，相同电流 I 流过每个电阻。总电势差 V 等于各电阻电势差之和：V = V₁ + V₂。利用欧姆定律，V = IR，V₁ = IR₁，V₂ = IR₂。
Substituting gives IR = IR₁ + IR₂. Dividing through by I yields the simple sum:
代入得 IR = IR₁ + IR₂。两边除以 I 得到简单的相加关系：
R = R₁ + R₂
This result can be extended to any number of resistors in series.
此结果可以推广到任意数量电阻串联的情形。
9. Deriving Total Resistance for Parallel Resistors (1/R = 1/R₁ + 1/R₂) | 推导并联电阻的总电阻 (1/R = 1/R₁ + 1/R₂)
. The total current I splits: I = I₁ + I₂. Using Ohm’s law, I = V/R, I₁ = V/R₁, I₂ = V/R₂.
在并联连接中，每个电阻两端电势差 V 相同。总电流 I 分流：I = I₁ + I₂。利用欧姆定律，I = V/R，I₁ = V/R₁，I₂ = V/R₂。
Therefore V/R = V/R₁ + V/R₂. Dividing by V gives the reciprocal formula:
因此 V/R = V/R₁ + V/R₂。除以 V 得到倒数公式：
1/R = 1/R₁ + 1/R₂
The combined resistance is always smaller than the smallest individual resistance.
并联后的总电阻总是小于最小的单个电阻。
10. Deriving the Wave Equation (v = fλ) | 推导波速方程 (v = fλ)
s per second. The period T is the time for one complete wave to pass a point, and f = 1/T.
波的频率 f 是每秒钟完整振动的次数。周期 T 是一个完整波通过某点的时间，且 f = 1/T。
tly one wavelength λ. Therefore speed v = distance/time = λ/T. Substituting T = 1/f gives:
在一个周期内，波恰好传播一个波长 λ。因此速度 v = 距离/时间 = λ/T。代入 T = 1/f 得到：
v = fλ
This fundamental relationship applies to all types of waves, from sound to electromagnetic waves.
这个基本关系适用于所有类型的波，从声波到电磁波。
Published by TutorHao | Physics Revision Series | aleveler.com
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