Constant acceleration formulae 1 | 匀加速度公式（一）

📚 Constant acceleration formulae 1 | 匀加速度公式（一）

In mechanics, when an object moves in a straight line with uniform acceleration, its motion can be described by a set of simple equations. These equations, often called the SUVAT equations, link the five fundamental quantities: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t). Mastering them is essential for solving a wide range of kinematics problems in A-Level Mathematics. This article introduces the five constant acceleration formulae, explains their derivation and use, and highlights common pitfalls. We will also explore motion under gravity as a key application.

在力学中，当物体沿直线做匀加速运动时，其运动规律可以用一组简洁的方程来描述。这些方程通常被称为 SUVAT 方程，它们将位移（s）、初速度（u）、末速度（v）、加速度（a）和时间（t）这五个基本量联系起来。掌握这些公式是解决 A-Level 数学中各类运动学问题的关键。本文将介绍五个匀加速度公式，解释它们的推导与用法，并指出常见误区。我们还将探讨重力作用下的运动作为重要的应用实例。

1. Introduction to the SUVAT Quantities | SUVAT 物理量简介

The five quantities used in constant acceleration problems are known by the acronym SUVAT: s for displacement measured in metres (m), u for initial velocity in metres per second (m·s⁻¹), v for final velocity in m·s⁻¹, a for constant acceleration in m·s⁻², and t for time in seconds (s). Displacement is a vector quantity, so its direction matters; the other quantities are also vectors in one-dimensional motion. You must choose a positive direction for the motion and stick to it consistently when assigning signs to u, v, a, and s.

匀加速问题中涉及的五个物理量常被缩写为 SUVAT：s 代表位移，单位为米（m）；u 为初速度，单位为米每秒（m·s⁻¹）；v 为末速度，单位也是 m·s⁻¹；a 是恒定的加速度，单位是米每平方秒（m·s⁻²）；t 为时间，单位是秒（s）。位移是矢量，因此方向至关重要；其他量在一维运动中同样也是矢量。解题时必须先规定一个正方向，并始终如一地对 u、v、a、s 赋予正负号。

Each SUVAT equation relates four of these five variables, so you can find the fifth if you know three others. This makes the system extremely useful for solving problems where the motion is uniformly accelerated, such as a car accelerating along a straight road or an object falling freely under gravity (ignoring air resistance). Remember that the acceleration must be constant – these equations do not work if the acceleration changes with time.

每个 SUVAT 方程联系了这五个量中的四个，因此只要已知其中三个，就能求出剩下的一个。这使得该方程组在求解匀加速运动问题时极为方便，例如汽车沿直路加速的运动，或物体在重力作用下忽略空气阻力的自由落体。务必牢记：只有加速度恒定不变时，这些公式才适用；若加速度随时间变化，则 SUVAT 方程不能直接使用。

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2. The Five Core Formulae | 五个核心公式

The five constant acceleration equations can be listed as follows. They all assume that acceleration a is constant, and motion takes place along a straight line. In the formulae, s represents the total displacement over the time interval t.

五个匀加速度公式列举如下。它们都假设加速度 a 恒定，且运动在直线上进行。公式中的 s 代表在时间间隔 t 内的总位移。

v = u + at

s = (u + v)t / 2

s = ut + ½ at²

s = vt − ½ at²

v² = u² + 2as

These equations are not independent; each can be derived from the definitions of velocity and acceleration, and from the velocity–time graph. You will often need to decide which formula to use based on which three quantities are known and which one is required. Practice in identifying the knowns and unknowns is key to selecting the most efficient equation.

这几个方程彼此并非独立；每一个都可以从速度与加速度的定义以及速度–时间图像中推导出来。解题时你需要根据已知的三个量和待求的一个量来选择最合适的公式。能快速识别已知与未知量，是高效选择方程的关键。

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3. Derivation from the Velocity–Time Graph | 由速度–时间图像推导

All five SUVAT equations can be derived from a straight-line velocity–time graph for constant acceleration. The line starts at u on the velocity axis and has gradient a, so after time t the velocity is v = u + at. The area under the graph represents the displacement s. By finding the area of a trapezium, we obtain s = (u + v)t / 2. By breaking the area into a rectangle and a triangle, we get s = ut + ½ at². Replacing u by v − at yields s = vt − ½ at². Finally, eliminating t from v = u + at and s = (u+v)t/2 gives v² = u² + 2as. Understanding these graphs deepens your insight into why the equations hold.

所有五个 SUVAT 公式都可以从匀加速情况下的速度–时间直线图像推出。该直线在速度轴上的截距为 u，斜率为加速度 a，所以经过时间 t 后的速度为 v = u + at。图像下方的面积代表位移 s。按梯形面积公式可得到 s = (u + v)t / 2；将面积分割为矩形和三角形，则得出 s = ut + ½ at²；用 v − at 替换 u 能导出 s = vt − ½ at²；最后，从 v = u + at 和 s = (u+v)t/2 中消去时间 t，便得到 v² = u² + 2as。深入理解这些图像能让你更透彻地掌握公式的来龙去脉。

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4. Using v = u + at | 使用 v = u + at

This is often the first equation used, since it links the velocities, acceleration and time. It is especially useful when displacement is not involved. For example: A car accelerates from rest at 3 m·s⁻² for 8 seconds. Find its final velocity. Here u = 0, a = 3, t = 8, so v = 0 + 3 × 8 = 24 m·s⁻¹. Always state the direction of the velocity if it matters; if the car reversed direction, you would assign a negative sign.

这通常是首先使用的公式，因为它直接关联了速度、加速度和时间，尤其适合不涉及位移的题目。例如：一辆汽车从静止出发，以 3 m·s⁻² 的加速度行驶 8 秒，求末速度。此处 u = 0，a = 3，t = 8，所以 v = 0 + 3 × 8 = 24 m·s⁻¹。若方向重要，务必标明速度的方向；如果车辆反向行驶，则需赋予负号。

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5. Using s = (u + v)t / 2 | 使用 s = (u + v)t / 2

This equation gives the displacement when the average velocity is multiplied by time. It is ideal when the initial and final velocities and time are known, and acceleration is not directly needed. Suppose a train slows down uniformly from 30 m·s⁻¹ to 10 m·s⁻¹ over 20 seconds. The displacement is s = (30 + 10) × 20 / 2 = 400 m. Notice that you do not need the acceleration to find s here.

该公式通过平均速度乘以时间得出位移。当已知初速度、末速度和时间，而不需要直接使用加速度时，它尤为方便。假设一列火车以匀减速从 30 m·s⁻¹ 降至 10 m·s⁻¹，历时 20 秒。位移为 s = (30 + 10) × 20 / 2 = 400 m。请注意，这里无需知道加速度即可求出 s。

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6. Using s = ut + ½ at² | 使用 s = ut + ½ at²

When acceleration is known but the final velocity is not, this formula is extremely useful. It often applies to objects starting with an initial velocity and undergoing uniform acceleration over a given time. For instance, a cyclist travels at 4 m·s⁻¹ and accelerates uniformly at 0.5 m·s⁻² for 10 seconds. The distance travelled is s = 4 × 10 + ½ × 0.5 × 10² = 40 + 25 = 65 m. Remember that s is displacement, not necessarily distance if direction changes; but in one-direction motion they coincide.

当加速度已知而未速度未知时，这个公式极为有用。它常应用于物体以某一初速度出发，经历给定时间匀加速运动的情况。例如，一名自行车手以 4 m·s⁻¹ 的速度行驶，并匀加速 0.5 m·s⁻² 持续 10 秒。所行位移为 s = 4 × 10 + ½ × 0.5 × 10² = 40 + 25 = 65 m。注意 s 是位移，如果方向改变，位移与路程未必相等；但在单向运动中，两者一致。

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7. Using s = vt − ½ at² | 使用 s = vt − ½ at²

This is the less commonly used counterpart of s = ut + ½ at². It expresses displacement in terms of the final velocity, acceleration and time. It can be handy when the final velocity is known rather than the initial velocity. For example, if a ball thrown upwards has a final velocity of 5 m·s⁻¹ downwards after 3 seconds, and acceleration due to gravity is −9.8 m·s⁻² (taking upward as positive), you might use this formula to find its displacement without first calculating u. Practise recognising which equation directly uses the data you have.

这是相对少用的对应公式，用末速度、加速度和时间来表达位移。当已知末速度而不知初速度时，它能派上用场。例如，一个向上抛出的球在 3 秒后具有 5 m·s⁻¹ 向下的末速度，重力加速度为 −9.8 m·s⁻²（取向上为正），你可以用此公式直接计算位移，而无需先求 u。要练习在解题时快速识别哪些数据可以直接套用哪个公式。

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8. Using v² = u² + 2as | 使用 v² = u² + 2as

When time is not mentioned in the question, this equation is your best choice. It relates the velocities, acceleration and displacement. A typical problem is finding the stopping distance of a vehicle. If a car moving at 20 m·s⁻¹ decelerates at 5 m·s⁻² (so a = −5), the distance to stop is found by setting v = 0: 0 = 20² + 2 × (−5) × s → 0 = 400 − 10s → s = 40 m. Always check that your calculated s makes sense physically.

当题目未提及时间时，此方程是最佳选择。它联系了速度、加速度和位移。一个典型问题是求车辆的刹车距离。若一辆汽车以 20 m·s⁻¹ 的速度行驶，并以 5 m·s⁻² 减速（因此 a = −5），则停车距离可由 v = 0 求得：0 = 20² + 2 × (−5) × s → 0 = 400 − 10s → s = 40 m。务必验证算出的 s 在物理上合理。

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9. Sign Conventions and Choosing a Positive Direction | 符号约定与正方向的选择

Choosing a consistent positive direction is the most crucial step in solving SUVAT problems correctly. Imagine a ball thrown vertically upwards from the ground with speed 12 m·s⁻¹. If we take upward as positive, then u = +12, a = −9.8 m·s⁻². When the ball falls back down, its velocity will become negative. If you choose downward as positive instead, the signs of u, v, a and s must all be adjusted accordingly. Without a clear convention, mistakes in signs lead to wrong displacement values or impossible times.

选准一个始终如一的正方向，是正确解出 SUVAT 问题的关键一步。设想一个球从地面以 12 m·s⁻¹ 的速度竖直上抛。若取向上为正，则 u = +12，a = −9.8 m·s⁻²。当球回落时，速度将变为负值。如果换成向下为正，那么 u、v、a 和 s 的符号必须相应调整。若不明确方向规定，符号错误就会导致位移值不对或出现不合理的负时间。

Many students forget that displacement, velocity and acceleration are vectors. For upward projection, the displacement when the ball returns to the ground is zero, not some positive value. The entire motion must be treated with the chosen sign convention from start to finish. A useful tip is to draw a quick diagram of the motion and label all vectors with their signs before writing any equation. This habit drastically reduces sign errors.

许多学生容易忘记位移、速度和加速度都是矢量。对于上抛运动，当球落回地面时，位移为零，而不是某个正值。整个运动自始至终都必须用所选的正负号约定来处理。一个实用的方法是：在写出任何方程之前，快速画一幅运动示意图，并标出所有矢量的正负号。这个习惯能极大地减少符号错误。

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10. Motion Under Gravity | 重力作用下的运动

One of the most common applications of constant acceleration formulae is vertical motion under gravity, where the acceleration has magnitude g = 9.8 m·s⁻² (or sometimes 9.81, or 10 for rough estimates) directed downwards. If you take upward as positive, a = −g. All the SUVAT equations then apply directly. For example, a stone dropped from rest from a cliff takes 3 seconds to hit the ground. With upward positive, u = 0, a = −9.8, t = 3, displacement s = ut + ½ at² = 0 + ½ × (−9.8) × 9 = −44.1 m. The negative sign means the stone ends up 44.1 m below the starting point, so the cliff height is 44.1 m.

匀加速度公式最常见的应用之一是在重力作用下的竖直运动，此时加速度的大小为 g = 9.8 m·s⁻²（有时用 9.81 或粗略取值 10），方向竖直向下。若取向上为正，则 a = −g。所有 SUVAT 方程可直接套用。例如，一块石头从悬崖边静止下落，3 秒后击中地面。取向上为正，u = 0，a = −9.8，t = 3，位移 s = ut + ½ at² = 0 + ½ × (−9.8) × 9 = −44.1 m。负号表示石头比起点低了 44.1 m，因此悬崖高度为 44.1 m。

In problems where an object is thrown upwards, the time to reach maximum height can be found by setting v = 0 in v = u + at. The total time of flight for a projectile that lands at the same vertical level is twice this value (ignoring air resistance). Understanding the symmetry of the motion can save time, but be careful to only apply symmetry when the launch and landing heights are equal. Always check the conditions before relying on shortcuts.

在物体上抛的问题中，到达最高点的时间可在 v = u + at 中令 v = 0 求得。对于落回同一水平面的抛体，忽略空气阻力时，总飞行时间是该时间的两倍。理解运动的对称性可以节省时间，但要留心只有抛出点与落地点等高时才可应用对称性。依赖捷径前务必检查条件。

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11. Common Mistakes and How to Avoid Them | 常见错误与避免方法

A frequent error is mixing up the formulae, such as using s = ut + ½ at² when v is needed. Always list the five symbols s, u, v, a, t and tick the ones given in the question and the one you need to find. Select the formula that omits the quantity you don’t need. Another error is forgetting that the equations apply only when acceleration is constant. If a problem involves two stages of motion with different accelerations, treat each stage separately with its own set of SUVAT equations, but ensure the final velocity of the first stage becomes the initial velocity of the next if they are connected.

一个常见错误是混淆公式，例如在需要 v 的时候用了 s = ut + ½ at²。解题时应始终列出 s、u、v、a、t 五个符号，标记题目给出的量和待求的量，然后选择恰好避开无关量的公式。另一个错误是忘记这些方程仅在加速度恒定时才成立。如果问题涉及两个不同加速度的运动阶段，则需分别对每个阶段使用独立的 SUVAT 方程组，并注意若两阶段相连，前一阶段的末速度将成为后一阶段的初速度。

Students also make sign errors by not treating downward velocities as negative when upward is positive, especially when both directions appear in the same question. For example, when a ball is thrown up and then falls past the point of release, its velocity changes sign. Remember that the signs of u, v, a, and s must all be consistent with your chosen positive direction. A quick sketch marking the positive axis and the directions of all vectors is invaluable.

学生们还常犯符号错误：当取向上为正时，却未将向下的速度视为负值，尤其在一个问题中出现两个方向时。例如球被向上抛出后又向下经过抛出点，速度的符号便发生改变。务必记住 u、v、a 和 s 的符号都必须与你选定的正方向一致。一张简图，标明正轴和所有矢量的方向，简直是价值连城的工具。

Finally, some problems give information like ‘a car accelerates from rest to 20 m/s over 200 m’ – here time is not given, so v² = u² + 2as is the natural choice. Train yourself to recognise the missing quantity. In SUVAT, one quantity is always absent from each equation; use that clue to pick the right one.

最后，有些题目给出如“一辆汽车从静止加速至 20 m/s，行驶了 200 m”这样的信息——此处时间未知，因此 v² = u² + 2as 是自然之选。要训练自己迅速识别缺失的量。在 SUVAT 中，每个公式总会缺少一个量；利用这一线索选出正确公式。

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12. Summary and Key Takeaways | 总结与要点

The five constant acceleration formulae provide a powerful toolkit for solving kinematics problems in one dimension with uniform acceleration. They rely on the interplay of displacement, initial velocity, final velocity, acceleration and time. Success depends on choosing a clear positive direction, identifying the known and unknown quantities, and selecting the appropriate formula. The velocity–time graph underpins all these equations, so drawing a graph can often clarify your thinking. Practice with a variety of scenarios, including freely falling bodies, decelerating vehicles, and two-stage motion, to build confidence. With these skills, you will be well-prepared for any straight-line constant acceleration question in Edexcel A-Level Mathematics.

五个匀加速度公式为求解一维匀加速运动学问题提供了强大的工具集。它们依赖于位移、初速度、末速度、加速度和时间之间的关联。成功的关键在于选定清晰的正方向，辨别已知量与未知量，并选用合适的公式。速度–时间图是所有公式的基础，因此绘制图像常常能理清思路。通过练习各种情境，包括自由落体、减速车辆和两段式运动，来建立信心。掌握这些技能后，你将能从容应对 Edexcel A-Level 数学中任何直线匀加速问题。

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