📚 Deriving the Equations of Motion | 运动学方程推导
In AS Physics, understanding how to derive the equations of motion for uniform acceleration is essential. These formulas, often called SUVAT equations, form the backbone of kinematics and appear frequently in exam questions, including those from the June 2019 series. Mastering their derivation not only helps you recall them under pressure but also deepens your grasp of how displacement, velocity, acceleration and time are linked.
在AS物理中，掌握匀加速运动方程的推导至关重要。这些公式常被称为SUVAT方程，是运动学的基础，经常出现在包括2019年6月试卷在内的考试题目中。掌握它们的推导不仅能帮助你在紧张时准确回忆，还能让你更深刻地理解位移、速度、加速度和时间之间的联系。
1. Understanding the Variables | 理解变量
Before deriving any equation, we must define the five key quantities. In the SUVAT set, s represents displacement (the straight-line distance from start to finish, with direction); u is the initial velocity at time t = 0; v is the final velocity after time t; a is the constant acceleration; and t is the elapsed time during which the motion occurs. All of these are measured in SI units: metres (m) for s, metres per second (m/s) for u and v, metres per second squared (m/s²) for a, and seconds (s) for t.
在推导任何方程之前，我们必须先定义五个关键物理量。在SUVAT体系中，s表示位移（从起点到终点的直线距离，具有方向）；u是t=0时的初速度；v是经过时间t后的末速度；a是恒定的加速度；t是运动经过的时间。这些量都使用国际单位：s用米（m），u和v用米每秒（m/s），a用米每二次方秒（m/s²），t用秒（s）。
The term ‘constant acceleration’ is crucial. It means that the rate of change of velocity does not vary. If acceleration were not constant, the standard SUVAT equations would no longer be valid. In many real-world situations, we approximate acceleration as constant, such as a car accelerating steadily on a straight road or an object falling freely under gravity when air resistance can be ignored.
“恒定加速度”这一条件是关键。它意味着速度的变化速率保持不变。如果加速度不是恒定的，标准SUVAT方程将不再成立。在许多实际情形中，我们将加速度近似为恒定，例如汽车在直路上平稳加速，或者忽略空气阻力时物体在重力作用下的自由下落。
2. The Definition of Acceleration | 加速度的定义
Acceleration is defined as the rate of change of velocity. In symbols: a = (v − u) / t, where (v − u) is the change in velocity Δv and t is the time taken for that change. Provided the acceleration is constant, we can rearrange this definition directly to obtain the first equation of motion.
加速度的定义是速度的变化率。用符号表示为：a = (v − u) / t，其中(v − u)是速度的变化量Δv，t是发生该变化所用的时间。只要加速度恒定不变，我们就可以直接重新整理这个定义式，得到第一个运动学方程。
Multiply both sides by t: a × t = v − u. Then add u to both sides to isolate v: v = u + at. This equation tells us that the final velocity v is simply the initial velocity plus the product of acceleration and time. It is often written as v = u + at. Be careful with signs: if the object is decelerating, the acceleration a is negative, and the equation still holds.
两边乘以t：a × t = v − u。然后将u加到等式两边以解出v：v = u + at。该方程告诉我们，末速度v等于初速度加上加速度与时间的乘积。它通常写成v = u + at。使用时要留意符号：如果物体在减速，加速度a取负值，方程依然成立。
3. Average Velocity and Displacement | 平均速度与位移
For an object moving with uniform acceleration, the velocity changes linearly with time. This means the average velocity over the interval is simply the arithmetic mean of the initial and final velocities: average velocity = (u + v) / 2. This statement is only true when the acceleration is constant, because the velocity-time graph is a straight line.
对于匀加速运动的物体，速度随时间呈线性变化。这意味着该时间段内的平均速度就是初速度与末速度的算术平均值：平均速度 = (u + v) / 2。这个结论仅在加速度恒定时成立，因为此时速度-时间图是一条直线。
Displacement s is the product of average velocity and total time t. Replacing average velocity with (u + v) / 2 gives us: s = ((u + v) / 2) × t, which is usually written as s = ½ (u + v) t. This equation links displacement to the initial and final velocities without directly involving acceleration.
位移s等于平均速度与总时间t的乘积。用(u + v)/2替换平均速度，我们得到：s = ((u + v) / 2) × t，通常记作s = ½ (u + v) t。这个方程将位移与初末速度联系起来，而不直接涉及加速度。
4. Eliminating Final Velocity | 消去末速度v
Sometimes we know the initial velocity, acceleration and time but not the final velocity. We can combine v = u + at with s = ½ (u + v) t. Substitute v from the first equation into the second: s = ½ (u + [u + at]) t = ½ (2u + at) t.
有些时候我们知道初速度、加速度和时间，但不知道末速度。我们可以将v = u + at与s = ½ (u + v) t联立。将第一个方程中的v代入第二个方程：s = ½ (u + [u + at]) t = ½ (2u + at) t。
Expand the bracket carefully: ½ × t × 2u = u t, and ½ × t × at = ½ a t². The result is s = u t + ½ a t². This is the most commonly used form for calculating displacement when initial velocity is known and acceleration is constant. Notice that the displacement consists of two terms: ‘ut’ is the distance the object would have travelled if there were no acceleration, and ‘½ a t²’ is the extra distance due to acceleration.
仔细展开括号：½ × t × 2u = u t，而 ½ × t × at = ½ a t²。结果是s = u t + ½ a t²。这是一个最常用的形式，用于已知初速度和恒定加速度时计算位移。请注意，位移由两项组成：“ut”是如果没有加速度时物体本该行进的距离，“½ a t²”则是加速度带来的额外距离。
5. Eliminating Time | 消去时间t
When time is not given or not asked for, it is useful to have an equation that directly connects velocity, acceleration and displacement. Starting again from v = u + at, we can solve for t: t = (v − u) / a.
当没有给出时间或不需要求时间时，有一个能直接联系速度、加速度和位移的方程会很有用。还是从v = u + at出发，我们可以解出t：t = (v − u) / a。
Now substitute this expression for t into s = ½ (u + v) t: s = ½ (u + v) × (v − u) / a. Simplify the numerator: (u + v)(v − u) = v² − u² (using the difference of two squares). Thus s = (v² − u²) / (2a).
现在将这个t的表达式代入s = ½ (u + v) t：s = ½ (u + v) × (v − u) / a。简化分子：(u + v)(v − u) = v² − u²（利用平方差公式）。于是得到s = (v² − u²) / (2a)。
Multiply both sides by 2a and rearrange: 2a s = v² − u², and then v² = u² + 2a s. This equation is extremely handy for problems involving take-off velocity, braking distances, or reaching a maximum height, where time is not a primary concern.
两边乘以2a并移项：2a s = v² − u²，继而得到v² = u² + 2a s。这个方程在涉及起飞速度、刹车距离或到达最大高度等问题中非常实用，因为这些情形下时间往往不是首要关注点。
6. Derivation Using a Velocity-Time Graph | 运用速度-时间图推导
All four SUVAT equations can be visualised and derived using a velocity-time (v‑t) graph. Draw a graph with velocity on the y-axis and time on the x-axis. For constant acceleration a, the line is straight, starting at (0, u) and ending at (t, v). The slope of the line is acceleration a = (v − u) / t, which immediately yields v = u + at.
所有四个SUVAT方程都可以利用速度-时间（v‑t）图进行直观推导。绘制一张图，纵轴为速度，横轴为时间。对于恒定加速度a，图线是一条直线，起点为(0, u)，终点为(t, v)。直线的斜率就是加速度a = (v − u) / t，这立刻得出v = u + at。
The area under the v‑t graph represents the displacement s. The area of this trapezium can be calculated as the average of the parallel sides multiplied by the base: area = ½ (u + v) × t. Hence s = ½ (u + v) t. Splitting the trapezium into a rectangle of area u × t and a triangle of area ½ (v − u) × t yields s = u t + ½ (v − u) t. Replacing (v − u) with a t gives s = u t + ½ a t². Eliminating t using the slope formula leads to v² = u² + 2a s. This geometric approach often helps students remember the equations more vividly.
v‑t图下方的面积代表位移s。这个梯形的面积等于平行边的平均值乘以底边长：面积 = ½ (u + v) × t。因此s = ½ (u + v) t。将梯形拆分为一个面积为u × t的矩形和一个面积为½ (v − u) × t的三角形，得到s = u t + ½ (v − u) t。再将(v − u)替换为a t就得到s = u t + ½ a t²。利用斜率公式消去t便可得v² = u² + 2a s。这种几何方法经常能帮助学生更形象地记住方程。
7. Derivation with Calculus (Extension) | 微积分推导（拓展）
For those studying A-level Mathematics alongside Physics, the same equations can be obtained elegantly through integration. Acceleration is the derivative of velocity with respect to time: a = dv/dt. If a is constant, we can integrate directly: ∫ dv = ∫ a dt → v = a t + C. The constant of integration C is determined by the initial condition: when t = 0, v = u, so C = u. Therefore v = u + a t.
对于同时修读A-level数学的物理学生来说，这些方程可以通过积分优雅地得到。加速度是速度对时间的导数：a = dv/dt。如果a恒定，我们可以直接积分：∫ dv = ∫ a dt → v = a t + C。积分常数C由初始条件确定：当t = 0时，v = u，所以C = u。因此v = u + a t。
Velocity itself is the derivative of displacement: v = ds/dt. Substituting the expression for v and integrating: s = ∫ (u + a t) dt = u t + ½ a t² + D. Again, taking the origin as the starting point, we set s = 0 at t = 0, which gives D = 0. Thus s = u t + ½ a t². The third SUVAT equation can then be derived by eliminating t from these two expressions, exactly as we did algebraically. This shows the deep consistency between calculus and kinematics.
速度本身又是位移的导数：v = ds/dt。代入v的表达式并进行积分：s = ∫ (u + a t) dt = u t + ½ a t² + D。同样，以出发点为坐标原点，我们设在t = 0时s = 0，这就得到D = 0。因此s = u t + ½ a t²。第三个SUVAT方程随后可通过消去t得到，完全和代数推导方法一致。这展示了微积分与运动学之间深刻的一致性。
8. Worked Example: Free Fall | 实例：自由落体
A classic application is an object dropped from rest near the Earth’s surface, where acceleration a = g = 9.81 m/s² downward. Take downward as positive. Given u = 0, we can find the displacement after 3.0 seconds using s = u t + ½ a t² = 0 + ½ × 9.81 × (3.0)² = 44.1 m (to 3 significant figures).
一个经典应用是物体在地表附近由静止自由下落，此时加速度a = g = 9.81 m/s²，方向向下。取向下为正方向。给定初速度u = 0，我们可以用s = u t + ½ a t²来计算3.0秒后的位移：s = 0 + ½ × 9.81 × (3.0)² = 44.1 m（保留三位有效数字）。
If we wanted the velocity just before hitting the ground from a known height of 20.0 m, we could use v² = u² + 2 a s. Plugging in u = 0, a = 9.81 m/s², s = 20.0 m: v² = 0 + 2 × 9.81 × 20.0 = 392.4, so v = √392.4 ≈ 19.8 m/s. The sign is positive because the motion is downward and we chose downward as positive. If instead we had taken upward as positive, we would set g = −9.81 m/s², and the answer would be negative, indicating velocity downwards.
如果我们想知道物体从20.0米高处落地前的速度，可以用v² = u² + 2 a s。代入u = 0，a = 9.81 m/s²，s = 20.0 m：v² = 0 + 2 × 9.81 × 20.0 = 392.4，所以v = √392.4 ≈ 19.8 m/s。速度符号为正，因为运动方向向下且我们规定向下为正。如果将向上取为正方向，则需设g = −9.81 m/s²，所得速度将为负值，表示速度向下。
9. Common Mistakes and Misconceptions | 常见错误与误解
One frequent error is using the equations when acceleration is not constant. For example, a car accelerating with a decreasing rate of pedal push does not follow a constant a. The SUVAT equations would not apply. Always check the problem statement for ‘constant acceleration’ or ‘uniform acceleration’.
一个常见错误是在加速度不恒定时使用这些方程。例如，一辆车以逐渐减小的油门加速，加速度并非恒定，此时SUVAT方程就不适用了。请务必检查题目中是否有“恒定加速度”或“匀加速”的字眼。
Another pitfall is sign convention. When solving problems involving vertical motion, decide early whether upward or downward is positive and stick to that choice for all vectors. A sign error in u, v, a or s can lead to physically impossible answers, such as a negative time. Also, do not write v = u + at as v = u + a/t; misplacing the multiplication is a surprisingly common slip.
另一个陷阱是符号规则。求解涉及竖直运动的问题时，要尽早确定向上还是向下为正，并对所有矢量统一使用这个选择。u、v、a或s的符号错误会导致物理上不合理的答案，比如时间为负。此外，不要把v = u + at误写成v = u + a/t；这种乘法误置是出人意料的常见笔误。
Students sometimes think that ‘s’ always represents total distance travelled. In SUVAT, s is displacement, which is a vector. If the object changes direction and returns past its start, the displacement could be zero while the distance is not. For problems involving distance, you may need to break the motion into sections and use the equations separately.
学生有时会认为“s”总是代表行驶的总路程。在SUVAT体系中，s是位移，是一个矢量。如果物体改变方向并回越起点，位移可能为零而路程却不为零。对于涉及路程的问题，你可能需要将运动分段并分别应用方程。
10. Summary and Key Points | 总结与要点
The four equations of motion for constant acceleration in a straight line are: v = u + at, s = ½ (u + v) t, s = u t + ½ a t², and v² = u² + 2 a s. Each equation omits one of the five variables: the first lacks s, the second lacks a, the third lacks v, and the fourth lacks t. Knowing which quantity is missing helps you choose the right equation efficiently during exams.
匀加速直线运动的四个运动学方程是：v = u + at，s = ½ (u + v) t，s = u t + ½ a t²，以及 v² = u² + 2 a s。每个方程都缺少五个变量中的一个：第一个缺少s，第二个缺少a，第三个缺少v，第四个缺少t。知道哪个物理量未出现可以帮助你在考试中高效地选择正确的方程。
Whether you derive them algebraically, geometrically or with calculus, the key assumptions are constant acceleration and motion along a straight line. Practise these derivations regularly – writing them out step by step – until they become second nature. This will not only boost your confidence in tackling mechanics questions but also prepare you for the deeper analysis required in A2 topics.
无论你使用代数、几何还是微积分来推导，关键假设都是恒定加速度和直线运动。定期练习这些推导——一步一步地写出来——直到它们成为你的第二天性。这不仅能增强你解答力学问题的信心，还能为你应对A2阶段更深入的分析做好准备。
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