📚 Constant Acceleration Formulae 1 | 恒定加速度公式 1
In Edexcel A-Level Mechanics, constant acceleration formulae are the core tools for describing motion in a straight line when acceleration does not change. They are often called the SUVAT equations because they link displacement, initial velocity, final velocity, acceleration and time.
在 Edexcel A-Level 力学中，恒定加速度公式是描述加速度不变时直线运动的核心工具。它们通常被称为 SUVAT 方程，因为它们把位移、初速度、末速度、加速度和时间联系起来。
This topic is essential for modelling cars, falling objects, projectiles and any motion where acceleration can be treated as uniform over a short interval. Mastering these equations is not only about memorising five formulas; it is about learning how to translate a physical situation into symbols, choose the correct equation, and interpret the answer with the correct direction and units.
这个主题对于建立汽车、落体、抛体以及任何加速度在短时间内可视为均匀的运动模型都至关重要。掌握这些方程不只是背下五个公式，而是要学会如何把物理情境转化为符号，选择正确的方程，并用正确的方向和单位解释答案。
1. What Is Constant Acceleration? | 什么是恒定加速度？
Constant acceleration means that the rate of change of velocity is fixed throughout the motion. In other words, the velocity changes by exactly the same amount in every equal time interval.
恒定加速度意味着速度的变化率在整个运动过程中保持不变。换句话说，在每个相等的时间间隔内，速度的变化量完全相同。
For example, if a car accelerates at 2 m s⁻² in a straight line, its velocity increases by 2 m s⁻¹ every second. This does not mean the velocity is constant; it means the acceleration is constant, so velocity changes linearly with time.
例如，如果一辆汽车沿直线以 2 m s⁻² 加速，它的速度每秒增加 2 m s⁻¹。这并不意味着速度恒定；它表示加速度恒定，因此速度随时间线性变化。
In A-Level mechanics, we usually restrict constant acceleration problems to motion in a straight line. This allows us to use positive and negative signs to show direction.
在 A-Level 力学中，我们通常将恒定加速度问题限制在直线运动。这样我们可以使用正负号来表示方向。
If acceleration is constant, the average velocity over an interval is simply the mean of the initial and final velocities, provided the direction does not reverse during that interval. This fact is the basis of one of the SUVAT equations.
如果加速度恒定，只要在这段时间内运动方向没有反向，区间内的平均速度就等于初速度与末速度的平均值。这一事实正是其中一个 SUVAT 方程的基础。
2. The Five SUVAT Variables | 五个 SUVAT 变量
= displacement, u = initial velocity, v = final velocity, a = acceleration, and t = time.
这五个变量最好用缩写 SUVAT 来记忆：s = 位移，u = 初速度，v = 末速度，a = 加速度，t = 时间。
Each SUVAT equation uses exactly four of these five variables. You must be able to identify which variable is not involved in a particular equation. This is the key skill in solving constant acceleration problems quickly and accurately.
每个 SUVAT 方程恰好使用这五个变量中的四个。你必须能够识别某个特定方程中未涉及哪个变量。这是快速且准确解决恒定加速度问题的关键技能。
| Quantity / 量 | Symbol / 符号 | SI Unit / 国际单位 | Typical Meaning / 典型含义 |
|---|---|---|---|
| Displacement / 位移 | s | m | Distance moved in a given direction / 沿指定方向移动的距离 |
| Initial velocity / 初速度 | u | m s⁻¹ | Velocity at the start of the interval / 区间开始时的速度 |
| Final velocity / 末速度 | v | m s⁻¹ | Velocity at the end of the interval / 区间结束时的速度 |
| Acceleration / 加速度 | a | m s⁻² | Rate of change of velocity / 速度的变化率 |
| Time / 时间 | t | s | Duration of the motion / 运动的持续时间 |
Note that displacement and velocity are vectors in one dimension, so their signs matter. Acceleration is also a vector. Time is always positive in these problems, so t is never negative.
注意，位移和速度在一维中是矢量，因此它们的正负号很重要。加速度也是矢量。在这类问题中时间始终为正，所以 t 永远不会是负数。
3. The Five SUVAT Equations | 五个 SUVAT 方程
There are five standard constant acceleration equations. Each one omits exactly one of the five SUVAT variables.
共有五个标准恒定加速度方程。每一个都恰好缺少五个 SUVAT 变量中的一个。
v = u + at
s = ½(u + v)t
s = ut + ½at²
s = vt − ½at²
v² = u² + 2as
e fourth, s = vt − ½at², does not contain u. The fifth, v² = u² + 2as, does not contain t.
第一个方程 v = u + at 不含 s。第二个 s = ½(u + v)t 不含 a。第三个 s = ut + ½at² 不含 v。第四个 s = vt − ½at² 不含 u。第五个 v² = u² + 2as 不含 t。
These equations are valid only when acceleration is constant and motion is in a straight line. If acceleration changes, you must split the motion into stages or use calculus.
这些方程仅在加速度恒定且运动为直线运动时成立。如果加速度变化，你必须把运动分成不同阶段或使用微积分。
4. Choosing the Right Equation | 如何选择正确的方程
To solve a SUVAT problem, list the known values, identify the unknown, and then choose the equation that contains those four quantities. The variable you do not know and do not need is the one that tells you which equation to use.
解决 SUVAT 问题时，先列出已知量，确定未知量，然后选择包含这四个量的方程。你不知道也不需要求的那个变量，就是决定使用哪个方程的线索。
o find a, use v² = u² + 2as. If you know u, v and t and want to find s, use s = ½(u + v)t.
例如，若已知 u、a、t，要求 s，就用 s = ut + ½at²。若已知 u、v、s，要求 a，就用 v² = u² + 2as。若已知 u、v、t，要求 s，就用 s = ½(u + v)t。
- Write down s, u, v, a, t in a column and fill in what you know. / 把 s、u、v、a、t 列成一列，填入已知量。
- Mark the unknown with a question mark. / 用问号标出未知量。
- Choose the equation that does not contain the variable you neither know nor need. / 选择不包含你既不知道也不需要求的变量的方程。
- Substitute numbers with units and solve. / 代入带单位的数值并求解。
- Check whether the answer is physically reasonable. / 检查答案在物理上是否合理。
5. Sign Conventions and Direction | 正负号与方向
In one-dimensional motion, you must choose a positive direction before substituting values. All vectors in the same direction as your positive direction are positive; all vectors in the opposite direction are negative.
在一维运动中，代入数值前必须先选定正方向。与正方向相同的所有矢量取正；与正方向相反的所有矢量取负。
If a car is moving to the right and slowing down, and you choose right as positive, then u is positive and a is negative. If you choose left as positive instead, then u is negative and a is positive. The physical answer is the same, but the signs in your working will be different.
如果汽车向右运动并减速，且你选择向右为正，则 u 为正，a 为负。如果你改选向左为正，则 u 为负，a 为正。物理答案相同，但计算过程中的符号会不同。
Displacement is not the same as distance. Displacement can be negative if the object ends up on the negative side of the starting point, while distance travelled is always positive.
位移与路程不同。如果物体最终位于起点的负方向一侧，位移可以为负，而走过的路程始终为正。
to gravity is a = −g, where g ≈ 9.8 m s⁻². If you take downwards as positive, then a = +g.
对于竖直运动，通常取向上为正。此时重力加速度为 a = −g，其中 g ≈ 9.8 m s⁻²。如果取向下为正，则 a = +g。
6. Worked Example 1: A Car Accelerating | 例题一：加速的汽车
A car starts from rest and accelerates uniformly at 2.5 m s⁻² for 8.0 s. Find its final velocity and the distance travelled.
一辆汽车从静止开始以 2.5 m s⁻² 匀加速运动 8.0 s。求它的末速度和行驶距离。
s⁻¹, a = 2.5 m s⁻², t = 8.0 s. Need v and s. Since s is not needed for v, use v = u + at.
已知：u = 0 m s⁻¹，a = 2.5 m s⁻²，t = 8.0 s。要求 v 和 s。求 v 时不需要 s，用 v = u + at。
v = 0 + 2.5 × 8.0 = 20 m s⁻¹
For distance, use s = ut + ½at². Since u = 0, the first term vanishes.
求距离时用 s = ut + ½at²。因为 u = 0，第一项为零。
s = 0 + ½ × 2.5 × 8.0² = 80 m
80 m. Notice that the units are consistent: m s⁻² × s gives m s⁻¹, and m s⁻² × s² gives m.
因此汽车达到 20 m s⁻¹，行驶 80 m。注意单位是一致的：m s⁻² × s 得到 m s⁻¹，m s⁻² × s² 得到 m。
7. Worked Example 2: Braking and Deceleration | 例题二：刹车与减速
A train is moving at 30 m s⁻¹ when the brakes are applied. It decelerates uniformly and stops in 15 s. Find the deceleration and the braking distance.
一列火车以 30 m s⁻¹ 行驶时开始刹车。它匀减速并在 15 s 内停下。求减速度与刹车距离。
of motion as positive. Then u = 30 m s⁻¹, v = 0 m s⁻¹, t = 15 s. Use v = u + at to find a.
取运动方向为正。则 u = 30 m s⁻¹，v = 0 m s⁻¹，t = 15 s。用 v = u + at 求 a。
0 = 30 + a × 15
a = −2.0 m s⁻²
e magnitude of the deceleration is 2.0 m s⁻². For the braking distance, use s = ½(u + v)t.
负号表示加速度与运动方向相反。减速度的大小为 2.0 m s⁻²。求刹车距离时用 s = ½(u + v)t。
s = ½ × (30 + 0) × 15 = 225 m
s gives 0 = 30² + 2 × (−2.0) × s, so s = 225 m. Both methods agree, which is a good check.
或者用 v² = u² + 2as：0 = 30² + 2 × (−2.0) × s，得 s = 225 m。两种方法一致，这是很好的检验。
8. Worked Example 3: Free Fall and g | 例题三：自由落体与 g
noring air resistance, find the time to reach the ground and the speed just before impact.
一个小球从 45 m 高处由静止释放。取 g = 9.8 m s⁻²，忽略空气阻力，求落地所需时间与落地前的速度。
Take downwards as positive. Then u = 0 m s⁻¹, a = 9.8 m s⁻², s = 45 m. Use s = ut + ½at².
取向下为正。则 u = 0 m s⁻¹，a = 9.8 m s⁻²，s = 45 m。用 s = ut + ½at²。
45 = 0 + ½ × 9.8 × t²
t² = 45 ÷ 4.9 ≈ 9.18
t ≈ 3.0 s
= 2 × 9.8 × 45, giving v ≈ 29.7 m s⁻¹. The positive sign means the velocity is downwards.
求落地速度可用 v = u + at = 0 + 9.8 × 3.03 ≈ 29.7 m s⁻¹，或用 v² = u² + 2as = 2 × 9.8 × 45，得 v ≈ 29.7 m s⁻¹。正号表示速度方向向下。
s is positive. SUVAT equations apply to each stage separately because the direction of motion changes.
如果小球改为向上抛出，运动就分为两个阶段。上升阶段 a = −g，下降阶段若取向下为正则 a = +g。由于运动方向改变，SUVAT 方程必须分别应用于每个阶段。
9. Graphs and Constant Acceleration | 图像与恒定加速度
On a velocity-time graph, constant acceleration is represented by a straight line. The gradient of the line equals the acceleration, and the area under the line equals the displacement.
在速度-时间图像上，恒定加速度表现为一条直线。直线的斜率等于加速度，直线下方的面积等于位移。
ly the second SUVAT equation. The gradient is (v − u) ÷ t, which rearranges to v = u + at.
对于从 u 到 v、历时 t 的恒定加速度，图像是一个梯形。其面积为 ½(u + v)t，正是第二个 SUVAT 方程。斜率为 (v − u) ÷ t，整理得 v = u + at。
On a displacement-time graph, constant acceleration produces a parabola because s depends on t². The gradient of the displacement-time graph gives the velocity at that instant.
在位移-时间图像上，恒定加速度产生抛物线，因为 s 与 t² 有关。位移-时间图像的斜率给出该时刻的速度。
- v-t graph: straight line, gradient = a, area = s. / 速度-时间图像：直线，斜率 = a，面积 = s。
- s-t graph: parabola, gradient = v. / 位移-时间图像：抛物线，斜率 = v。
- Negative gradient on a v-t graph means deceleration relative to the chosen positive direction. / 速度-时间图像上的负斜率表示相对于所选正方向的减速。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
Many marks are lost not because the algebra is difficult, but because the problem is set up carelessly. The most common mistake is mixing units, such as using km h⁻¹ with seconds without converting.
很多失分并不是因为代数难，而是因为题目设置不严谨。最常见的错误是单位混用，例如把 km h⁻¹ 与秒一起使用却没有换算。
Writing a = 2 m s⁻² for a braking car is a sign error that changes the answer completely.
另一个常见错误是忘记：当运动方向为正时，减速意味着加速度为负。对刹车中的汽车写成 a = 2 m s⁻² 是符号错误，会彻底改变答案。
Always check whether the object changes direction during the motion. If it does, the SUVAT equations can still be used, but the displacement and velocity signs must be handled carefully, and you may need to split the motion.
始终检查物体在运动过程中是否改变方向。如果改变，SUVAT 方程仍然可以使用，但必须小心处理位移与速度的正负号，并且可能需要把运动分段。
- Convert km h⁻¹ to m s⁻¹ by dividing by 3.6. / 将 km h⁻¹ 除以 3.6 换算为 m s⁻¹。
- Do not confuse s (displacement) with total distance. / 不要把 s（位移）与总路程混淆。
- Use g = 9.8 m s⁻² unless the question says otherwise. / 除非题目另有说明，使用 g = 9.8 m s⁻²。
- Write the equation before substituting numbers. / 先写方程，再代入数值。
- Give answers with units and, where relevant, direction. / 答案要带单位，必要时说明方向。
11. Brief Derivation of the SUVAT Equations | SUVAT 方程的简要推导
The equations can be derived from the definitions of acceleration and average velocity. Acceleration is the rate of change of velocity:
这些方程可以从加速度与平均速度的定义推导出来。加速度是速度的变化率：
a = (v − u) ÷ t
ant acceleration, average velocity is ½(u + v), and displacement equals average velocity multiplied by time:
整理得 v = u + at。对于恒定加速度，平均速度为 ½(u + v)，位移等于平均速度乘以时间：
s = ½(u + v)t
ves v² = u² + 2as. The fourth equation, s = vt − ½at², follows by substituting u = v − at.
把 v = u + at 代入该式得 s = ut + ½at²。从 v = u + at 与 s = ½(u + v)t 中消去 t，得 v² = u² + 2as。第四个方程 s = vt − ½at² 可通过代入 u = v − at 得到。
Understanding where the equations come from helps you remember them and helps you see why they fail when acceleration is not constant.
理解这些方程的来源有助于记忆，也有助于明白为什么当加速度不恒定时它们不再适用。
12. Practice Questions | 练习题
Try these questions after reviewing the notes above. They cover the main skills: identifying variables, choosing an equation, handling signs and interpreting results.
复习以上内容后，尝试以下题目。它们涵盖主要技能：识别变量、选择方程、处理正负号以及解释结果。
- A cyclist accelerates uniformly from 4 m s⁻¹ to 10 m s⁻¹ in 3 s. Find the acceleration and distance travelled. / 自行车手在 3 s 内从 4 m s⁻¹ 匀加速到 10 m s⁻¹。求加速度与行驶距离。
- at 14 m s⁻¹. Taking g = 9.8 m s⁻², find the maximum height and the total time of flight. / 石块以 14 m s⁻¹ 竖直向上抛出。取 g = 9.8 m s⁻²，求最大高度与总飞行时间。
- A plane lands at 70 m s⁻¹ and decelerates uniformly at 3.5 m s⁻². Find the time and distance required to stop. / 飞机以 70 m s⁻¹ 着陆，并以 3.5 m s⁻² 匀减速。求停下所需时间与距离。
- A particle has initial velocity 5 m s⁻¹ and acceleration −1.5 m s⁻². Find its velocity and displacement after 6 s. / 质点初速度 5 m s⁻¹，加速度 −1.5 m s⁻²。求 6 s 后的速度与位移。
For each question, begin by writing the five SUVAT variables and marking the known and unknown quantities. This habit will make even unfamiliar problems feel manageable.
对每个问题，先写出五个 SUVAT 变量，标出已知量与未知量。这个习惯会让即使不熟悉的问题也变得可控。
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