📚 Constant Acceleration | 匀加速度运动
In Edexcel A-Level Mechanics, constant acceleration problems form the foundation of kinematics. These problems assume that acceleration does not change with time, so the standard SUVAT equations can be used to connect displacement, initial velocity, final velocity, acceleration and time. This article builds the topic up from the definition, through the five equations and their graph interpretation, to vertical motion, multi-stage motion, two-particle problems and the exam technique that turns a messy worded question into a clean calculation.
在 Edexcel A-Level 力学中，匀加速度问题是运动学的基础。这类问题假设加速度不随时间变化，因此可以用标准 SUVAT 公式把位移、初速度、末速度、加速度和时间联系起来。本文从定义出发，依次讲解五个公式及其图像意义，再延伸到竖直运动、多阶段运动、双物体问题，以及把繁琐文字题转化为清晰计算所需的应试技巧。
1. What is Constant Acceleration? | 什么是匀加速度？
Acceleration is defined as the rate of change of velocity, so constant acceleration means that the velocity changes by the same amount in every equal interval of time. In symbols, the average acceleration over an interval is the change in velocity divided by the time taken, and if that value is the same for every interval, the acceleration is constant.
加速度定义为速度的变化率，因此匀加速度意味着在每一段相等的时间间隔内，速度的变化量相同。用符号表示，某段时间内的平均加速度等于速度变化量除以所用时间；如果每一段时间算出的值都相同，加速度就是恒定的。
In one-dimensional motion, acceleration is a vector quantity with both magnitude and direction. In Edexcel mechanics the symbol a is measured in metres per second squared, written m/s², and it can be positive or negative depending on which direction you have chosen as positive. A negative value does not automatically mean "slowing down"; it means the acceleration points in the negative direction.
在一维运动中，加速度是既有大小又有方向的矢量。在 Edexcel 力学中，符号 a 的单位是米每二次方秒，写作 m/s²；它的正负取决于你选定的正方向。负值并不自动等于"减速"，它只表示加速度指向负方向。
For constant acceleration, the velocity-time graph is a straight line. The gradient of this line equals the acceleration, and the area between the line and the time axis equals the displacement. These two facts are the reason the SUVAT equations exist at all, and quoting them in a written answer often earns method marks.
对于匀加速度，速度-时间图像是一条直线。这条直线的斜率等于加速度，直线与时间轴之间的面积等于位移。这两点正是 SUVAT 公式存在的根本原因，在书面作答中写出这一理解往往能拿到方法分。
The five key variables used in the SUVAT equations are u (initial velocity), v (final velocity), a (constant acceleration), s (displacement) and t (time). Every constant acceleration question you meet at A-Level will give you three of these and ask for a fourth; the fifth is simply ignored.
SUVAT 公式中使用的五个关键变量是 u（初速度）、v（末速度）、a（匀加速度）、s（位移）和 t（时间）。A-Level 阶段遇到的每一道匀加速度题，本质上都是给出其中三个量、要求第四个量，剩下的那一个则被忽略。
- Assumption 1: acceleration is constant in both magnitude and direction. 假设一：加速度的大小和方向都保持不变。
- Assumption 2: motion is in a straight line. 假设二：运动沿直线进行。
- Assumption 3: air resistance and other resistive forces are negligible unless stated otherwise. 假设三：除题目另有说明，空气阻力等阻力可忽略。
2. The SUVAT Equations | SUVAT 公式
When acceleration is constant, the following five equations link the five variables. They are often called the SUVAT equations in the UK specification, and they are all listed in the formula booklet, so you are never expected to memorise them blindly — but you are expected to use them fluently.
当加速度恒定时，下面五个公式把这五个变量联系起来。在英国考纲中它们通常被称为 SUVAT 公式，且全部列在公式表中，因此并不要求死记硬背，但必须做到运用熟练。
v = u + at
s = ut + ½at²
s = vt − ½at²
v² = u² + 2as
s = ((u + v)/2) × t
hird come from the area under a velocity-time graph, split into a rectangle plus a triangle (or a rectangle minus a triangle). The fourth is obtained by eliminating t between the first two, and the fifth is simply the area of a trapezium, or equivalently the average velocity multiplied by the time.
第一个公式直接来自加速度的定义：把 a = (v − u)/t 变形即可。第二、第三式来自速度-时间图像下的面积，把面积拆成一个矩形加一个三角形（或一个矩形减一个三角形）。第四式由前两式消去 t 得到，第五式则就是梯形面积，等价于平均速度乘以时间。
| Equation 公式 | Missing 缺少 | Typical use 典型用途 |
|---|---|---|
| v = u + at | s | Finding speed after a known time 求经过已知时间后的速度 |
| s = ut + ½at² | v | Finding distance travelled in a known time 求已知时间内的路程 |
| s = vt − ½at² | u | When final velocity is known instead of initial 已知末速度而非初速度时 |
| v² = u² + 2as | t | When time is not given and not asked for 未给时间也不求时间时 |
| s = ((u + v)/2) × t | a | When acceleration is unknown 加速度未知时 |
Notice that each equation omits exactly one variable. That single observation is the whole strategy for selecting the right equation, and it is discussed in the next section.
注意每个公式恰好缺少一个变量。仅凭这一点，就足以构成选择正确公式的完整策略，下一节将详细讨论。
3. Choosing the Right Equation | 如何选择合适的公式
The standard method has four steps. First, sketch the situation and mark the positive direction. Second, list the values of u, v, a, s and t that you know, writing a question mark beside the unknown you want. Third, cross out the variable that appears neither in your data nor in the question. Fourth, pick the equation in which that crossed-out variable does not appear.
标准方法分四步。第一，画出情境图并标明正方向。第二，列出已知的 u、v、a、s、t 数值，并在所求未知量旁写上问号。第三，划掉既不在已知数据中、也不是题目所问的那个变量。第四，选择不包含该被划掉变量的公式。
s required; time t is not mentioned, so you use v² = u² + 2as, giving v² = 64 + 36 = 100 and v = 10 m/s. No unnecessary quadratic equation appears, which is exactly the point of the method.
例如，一名骑行者以 8 m/s 行驶，以 1.2 m/s² 匀加速前进 15 m，要求末速度。这里 u = 8，a = 1.2，s = 15，所求为 v；时间 t 未被提及，所以用 v² = u² + 2as，得 v² = 64 + 36 = 100，v = 10 m/s。全程没有出现多余的二次方程，这正是该方法的要义。
of the five quantities are unknown but one of them can be deduced from the geometry of the problem, such as "comes to rest" meaning v = 0 or "starts from rest" meaning u = 0, do that deduction first.
若题目给出时间、要求路程，则用 s = ut + ½at²。若给出时间、要求末速度，则用 v = u + at。如果五个量中有两个未知，但其中之一可由题目情境推出，例如"最终停止"意味着 v = 0，"由静止开始"意味着 u = 0，那么应先完成这类推导。
- "Starts from rest" → u = 0. "由静止开始" → u = 0。
- "Comes to rest" / "stops" → v = 0. "最终静止"/"停下" → v = 0。
- "Maximum height" → v = 0 at that instant. "最高点" → 该时刻 v = 0。
- "Returns to the starting point" → s = 0. "回到出发点" → s = 0。
- "Travels the last 20 m" → treat as a fresh stage with its own u. "最后 20 m" → 作为一个新阶段处理，另设初速度。
4. Sign Conventions and Vectors | 正负号约定与矢量
Sign errors are the single largest source of lost marks in this topic. Before writing any equation, choose one direction as positive and state it clearly. All quantities measured in that direction are then positive; all quantities measured in the opposite direction are negative.
正负号错误是本主题失分的最大来源。在写出任何公式之前，先选定一个方向为正，并明确写出。凡沿该方向测量的量取正，凡沿相反方向测量的量取负。
n upwards has a = −g, and an object falling has a = +g if downwards is positive. Both choices are equally valid; only consistency matters.
常见约定是取初始运动方向为正。此时减速物体的加速度为负；向上抛出的物体 a = −g；若取向下为正，则下落物体 a = +g。两种取法同样有效，唯一要求是前后一致。
Remember also that displacement s is a vector, not the total distance travelled. A particle that travels 10 m forward and then 4 m back has a displacement of +6 m but a distance of 14 m. If a question asks for total distance, you must split the motion into stages, find the distance in each stage, and add the magnitudes.
还要记住，位移 s 是矢量，不是走过的总路程。一个质点先向前 10 m、再后退 4 m，位移是 +6 m，而路程是 14 m。若题目要求总路程，必须把运动分段，求出每一段的路程，再把大小相加。
s the question states otherwise. Some textbooks use 9.81 m/s²; check the front of the paper or the instruction in the question, and quote the value you used in your working so the examiner can follow your arithmetic.
在 Edexcel 试卷中，除非题目另有说明，重力加速度取 g = 9.8 m/s²。部分教材使用 9.81 m/s²；请查看试卷卷首说明或题目提示，并在解题过程中写出你采用的值，以便考官核对你的计算。
5. Velocity-Time Graphs | 速度-时间图像
For constant acceleration the velocity-time graph is a straight line, so the mathematics of the topic and the geometry of the graph are two views of the same thing. The gradient gives the acceleration, and the signed area between the line and the time axis gives the displacement.
匀加速度下，速度-时间图像是一条直线，因此本主题的数学与图像几何是同一件事的两种视角。斜率给出加速度，直线与时间轴之间的有向面积给出位移。
If the line lies below the time axis, the velocity is negative and the area counts as negative displacement — the object is moving backwards relative to your chosen positive direction. If the line crosses the time axis, the object reverses direction at that instant, and total distance must be found by treating the two triangles separately and adding their magnitudes.
若直线位于时间轴下方，则速度为负，面积记为负位移——物体正沿与所选正方向相反的方向运动。若直线穿过时间轴，物体在该时刻反向；此时总路程必须把两个三角形分开计算，再把大小相加。
On an acceleration-time graph, constant acceleration appears as a horizontal line, and the area under it gives the change in velocity. On a displacement-time graph, the gradient gives the velocity, and constant acceleration produces a parabola (a quadratic curve). Being able to move between the three graphs is a frequently examined skill.
在加速度-时间图像上，匀加速度表现为一条水平线，其下方面积给出速度变化量。在位移-时间图像上，斜率给出速度，匀加速度对应一条抛物线（二次曲线）。能在三种图像之间自由转换，是高频考查的能力。
- s-t graph: gradient = velocity, curvature = acceleration. s-t 图：斜率 = 速度，弯曲程度反映加速度。
- v-t graph: gradient = acceleration, area = displacement. v-t 图：斜率 = 加速度，面积 = 位移。
- a-t graph: height = acceleration, area = change in velocity. a-t 图：纵坐标 = 加速度，面积 = 速度变化量。
6. Vertical Motion Under Gravity | 竖直方向的重力运动
Motion under gravity is the most common application of constant acceleration in Edexcel Mechanics, because near the Earth's surface the acceleration of a freely falling object is constant at about 9.8 m/s² downwards, provided air resistance is ignored.
重力作用下的运动是 Edexcel 力学中最常见的匀加速度应用，因为在忽略空气阻力的前提下，地表附近自由落体的加速度恒为约 9.8 m/s²，方向向下。
m/s² for the entire flight, whether the object is rising or falling, and s becomes negative if the object ends up below its starting point.
选定正方向后，整道题都要沿用。若取向上为正，则整个飞行过程中 a = −9.8 m/s²，无论物体上升还是下落都如此；若物体最终落在出发点下方，则 s 为负值。
o maximum height from v = u + at. Second, because acceleration is constant, the motion is symmetric: the time up equals the time down, and the speed on returning to the launch height equals the launch speed.
有两条结论能让竖直运动题快速求解。第一，在最高点速度瞬时为零，因此由 v = u + at 令 v = 0，即可求得上升到最高点的时间。第二，由于加速度恒定，运动具有对称性：上升时间等于下降时间，回到抛出高度时的速率等于抛出速率。
point. The maximum height is s = ut + ½at² = 19.6 × 2 − ½ × 9.8 × 4 = 39.2 − 19.6 = 19.6 m, and the ball returns to the ground 4.0 s after launch.
例如，一个球以 19.6 m/s 竖直向上抛出。取向上为正，a = −9.8 m/s²，由 v = 0 得上升到最高点的时间 t = 19.6 ÷ 9.8 = 2.0 s。最大高度 s = ut + ½at² = 19.6 × 2 − ½ × 9.8 × 4 = 39.2 − 19.6 = 19.6 m，球在抛出后 4.0 s 回到地面。
= 0 (the launch instant) or t = 4.0 s, which is the landing time. Recognising that s = 0 shortcuts a great deal of work.
注意，从抛出到落回地面的整个过程中位移为零，因此在 s = ut + ½at² 中取 s = 0。解 0 = 19.6t − 4.9t² 得 t = 0（抛出瞬间）或 t = 4.0 s，后者即落地时间。识别出 s = 0 可以省去大量计算。
7. Worked Example 1: Single Stage | 例题一：单阶段运动
A car starts from rest and accelerates uniformly at 2.5 m/s² for 6.0 s. Find (a) its final velocity and (b) the distance travelled. Take the direction of motion as positive.
一辆汽车由静止开始，以 2.5 m/s² 匀加速运动 6.0 s。求 (a) 末速度；(b) 行驶距离。取运动方向为正。
Step 1 — list the data: u = 0, a = 2.5 m/s², t = 6.0 s, v = ?, s = ? equested, so we need two equations; both are available because we have three known values.
第一步——列出数据：u = 0，a = 2.5 m/s²，t = 6.0 s，v = ?，s = ?。题目要求两个未知量，需要两个方程；由于已有三个已知量，两个方程都可用。
Step 2 — find v using v = u + at = 0 + 2.5 × 6.0 = 15 m/s.
第二步——用 v = u + at 求 v：v = 0 + 2.5 × 6.0 = 15 m/s。
Step 3 — find s using s = ut + ½at² = 0 + ½ × 2.5 × 36 = 45 m.
第三步——用 s = ut + ½at² 求 s：s = 0 + ½ × 2.5 × 36 = 45 m。
Step 4 — eck with the equation you did not use; it costs ten seconds and catches most arithmetic slips.
第四步——用另一个独立公式检验：v² = u² + 2as 给出 225 = 0 + 2 × 2.5 × s，得 s = 45 m，结果一致。养成用未使用的公式复核的习惯，只需十秒，却能发现绝大多数算术错误。
8. Worked Example 2: Multi-Stage Motion | 例题二：多阶段运动
A train accelerates uniformly from rest at 2.0 m/s² until it reaches 16 m/s. It then travels at this constant speed for 10 s, and finally brakes uniformly to rest in 5.0 s. Find the total distance travelled and the average speed for the whole journey.
一列火车由静止以 2.0 m/s² 匀加速，直到速度达到 16 m/s；随后以此速度匀速行驶 10 s；最后在 5.0 s 内匀减速至停止。求全程总距离与平均速率。
Stage 1 — as: 256 = 0 + 4s, so s₁ = 64 m. The time is t = 16 ÷ 2.0 = 8.0 s, which we may need later.
阶段一——加速。u = 0，v = 16，a = 2.0。由 v² = u² + 2as 得 256 = 0 + 4s，故 s₁ = 64 m。所用时间 t = 16 ÷ 2.0 = 8.0 s，后面可能用到。
Stage 2 — nt speed. Here a = 0, so the SUVAT equations still apply with a = 0. s₂ = 16 × 10 = 160 m.
阶段二——匀速。此处 a = 0，公式依然适用，只是代入 a = 0。s₂ = 16 × 10 = 160 m。
Stage 3 — and a = (0 − 16) ÷ 5.0 = −3.2 m/s². Then s₃ = ((u + v)/2) × t = (16 + 0) ÷ 2 × 5.0 = 40 m.
阶段三——制动。u = 16，v = 0，t = 5.0，a = (0 − 16) ÷ 5.0 = −3.2 m/s²。于是 s₃ = ((u + v)/2) × t = (16 + 0) ÷ 2 × 5.0 = 40 m。
= 64 + 160 + 40 = 264 m. Total time = 8.0 + 10 + 5.0 = 23 s. Average speed = 264 ÷ 23 = 11
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