Constant Acceleration Formulae | 匀加速运动公式

📚 Constant Acceleration Formulae | 匀加速运动公式

Constant acceleration formulae, commonly known as SUVAT equations, are fundamental in A-Level Mechanics. They describe the motion of a particle moving in a straight line with uniform acceleration. Mastering these equations allows you to solve a wide range of kinematics problems involving displacement, velocity, acceleration, and time.

匀加速运动公式，通常称为SUVAT方程，是A-Level力学中的基础。它们描述了一个质点沿直线以恒定加速度运动的情况。掌握这些公式可以帮助你解决涉及位移、速度、加速度和时间的各种运动学问题。

1. Introduction to Constant Acceleration | 匀加速运动简介

In mechanics, 'constant acceleration' means the rate of change of velocity is uniform and does not vary with time. This situation occurs when a net constant force acts on an object, such as gravity near the Earth's surface (ignoring air resistance). The motion is always along a straight line.

在力学中，“匀加速度”意味着速度的变化率是恒定的，不随时间变化。这种情况发生在一个恒定的净力作用于物体时，例如在地球表面附近的重力（忽略空气阻力）。运动总是沿直线进行。

Under constant acceleration, we can derive a set of equations that link five key quantities: displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t). These equations are only valid when acceleration is constant.

在匀加速条件下，我们可以推导出一组联系五个关键量的方程：位移(s)、初速度(u)、末速度(v)、加速度(a)和时间(t)。这些方程仅在加速度恒定时有效。

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2. The Five SUVAT Variables | 五个SUVAT变量

Before using the equations, it is essential to understand the five variables and their standard symbols and units:

在使用方程之前，必须理解这五个变量及其标准符号和单位：

• s – displacement (m) – a vector quantity; can be positive or negative depending on direction.

• s – 位移（米）– 矢量；根据方向可为正或负。

• u – initial velocity (m s⁻¹) – velocity at the start of the time interval.

• u – 初速度（米/秒）– 时间间隔开始时的速度。

• v – final velocity (m s⁻¹) – velocity at the end of the time interval.

• v – 末速度（米/秒）– 时间间隔结束时的速度。

• a – acceleration (m s⁻²) – constant rate of change of velocity.

• a – 加速度（米/秒²）– 速度恒定的变化率。

• t – time (s) – the duration of the motion being considered.

• t – 时间（秒）– 所考虑运动的持续时间。

Note: In SUVAT problems, u, v, a, and s are vectors along a straight line, so sign conventions (positive direction) must be clearly defined.

注意：在SUVAT问题中，u、v、a和s是沿直线的矢量，因此必须明确定义符号惯例（正方向）。

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3. Derivation of the Formulae | 公式推导

ith the definition of acceleration: a = (v − u) / t. Rearranging gives the first equation.

SUVAT方程可以从速度和加速度的定义推导出来，假设a恒定。从加速度的定义开始：a = (v − u) / t。移项得到第一个公式。

y is (u + v)/2. Displacement equals average velocity × time, which gives s = (u + v)t / 2.

此外，由于速度均匀变化，平均速度为 (u + v)/2。位移等于平均速度 × 时间，因此得到 s = (u + v)t / 2。

Combining these expressions yields the other equations. We will present each formula with its derivation and use.

结合这些表达式可以推出其他公式。我们将展示每个公式及其推导和用途。

These derivations highlight the importance of integration and the area under a velocity–time graph, which is a powerful visual tool.

这些推导突出了积分和速度-时间图下面积的重要性，这是一种强大的直观工具。

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4. Equation 1: v = u + at | 公式一：v = u + at

ost direct consequence of the definition of acceleration. Rearranging a = (v − u)/t gives:

这是加速度定义最直接的推论。将 a = (v − u)/t 移项得：

v = u + at

It relates final velocity, initial velocity, acceleration, and time. It is used when time is known or required, but displacement is not involved.

它关联了末速度、初速度、加速度和时间。当已知或需要时间但不涉及位移时使用。

Example: s⁻¹ to 25 m s⁻¹ in 4 s. Find acceleration. Using v = u + at → 25 = 5 + a×4 → a = 5 m s⁻².

示例：一辆汽车在4秒内从5米/秒加速到25米/秒。求加速度。使用 v = u + at → 25 = 5 + a×4 → a = 5 米/秒²。

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5. Equation 2: s = (u + v)t / 2 | 公式二：s = (u + v)t / 2

rly, so average velocity = (u + v)/2. Displacement is average velocity multiplied by time:

由于加速度恒定，速度线性变化，因此平均速度 = (u + v)/2。位移是平均速度乘以时间：

s = ½ (u + v) t

This formula does not contain acceleration, making it useful when a is unknown.

这个公式不包含加速度，因此在a未知时很有用。

Example: s uniformly from 2 m s⁻¹ to 8 m s⁻¹ over 10 s. Find displacement. s = (2+8)/2 × 10 = 50 m.

示例：一个质点从2米/秒匀加速到8米/秒，用时10秒。求位移。 s = (2+8)/2 × 10 = 50米。

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6. Equation 3: s = ut + ½at² | 公式三：s = ut + ½at²

s equation relates displacement, initial velocity, acceleration and time, but does not involve final velocity v.

将 v = u + at 代入 s = (u+v)t/2 得到 s = (u + u + at)t/2 = ut + ½at²。这个方程关联了位移、初速度、加速度和时间，但不涉及末速度v。

s = ut + ½ a t²

It is especially useful for motion starting from rest (u = 0), where s = ½at².

它对于从静止开始运动（u = 0）特别有用，此时 s = ½at²。

Example: A stone falls from rest under gravity (a = 9.8 m s⁻²). How far does it fall in 3 seconds? u=0 → s = 0 + ½×9.8×3² = 44.1 m.

示例：一块石头从静止自由下落，重力加速度 a = 9.8 米/秒²。3秒内下落多远？ u=0 → s = 0 + ½×9.8×3² = 44.1 米。

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7. Equation 4: s = vt − ½at² | 公式四：s = vt − ½at²

= ut + ½at², yielding s = (v − at)t + ½at² = vt − ½at². It eliminates initial velocity u.

这个公式较少被记忆，但可以通过将 u = v − at 代入 s = ut + ½at² 推导得到 s = (v − at)t + ½at² = vt − ½at²。它消去了初速度u。

s = vt − ½ a t²

It is handy when final velocity v is known and u is not needed. Some exam boards treat this as a variation of the standard set.

当已知末速度v而不需要u时，这个公式很方便。一些考试局将其视为标准方程组的一个变体。

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8. Equation 5: v² = u² + 2as | 公式五：v² = u² + 2as

using s = ut + ½at² and substituting t = (v−u)/a, we obtain the time-independent equation:

通过消去 v = u + at 和 s = (u+v)t/2 中的时间t，或利用 s = ut + ½at² 并代入 t = (v−u)/a，我们得到不含时间的方程：

v² = u² + 2 a s

This is extremely useful when time is not given or asked for. Note that s is the displacement, not distance, so signs matter.

当未给出或不求时间时，这个方程极为有用。注意 s 是位移，不是路程，因此符号很重要。

Example: long. Find deceleration. 0² = 20² + 2a×40 → a = −400/80 = −5 m s⁻² (deceleration 5 m s⁻²).

示例：一辆汽车以20米/秒的速度刹车滑行至停止(v=0)。滑痕长40米。求减速度。0² = 20² +

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