Constant Acceleration Formulae | 匀加速运动公式

📚 Constant Acceleration Formulae | 匀加速运动公式

Constant acceleration formulae, often called the SUVAT equations, are essential for solving linear motion problems in Edexcel A Level Mechanics. They connect displacement, initial velocity, final velocity, acceleration, and time when acceleration does not change.

匀加速运动公式，通常称为 SUVAT 方程，是解决 Edexcel A Level 力学中直线运动问题的核心工具。当加速度保持不变时，这些公式将位移、初速度、末速度、加速度和时间联系起来。

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1. The SUVAT Variables and Conditions | SUVAT 变量与适用条件

The five SUVAT variables are used to describe motion in a straight line. The equations are only valid when acceleration is constant for the entire period being considered, or for one stage of a multi-stage motion.

五个 SUVAT 变量用于描述直线运动。这些公式仅在所考虑的整个时间段内加速度保持不变，或仅适用于多阶段运动中的某一个阶段。

Symbol	Meaning | 含义	SI Unit | 国际单位
s	Displacement | 位移	m
u	Initial velocity | 初速度	m s⁻¹
v	Final velocity | 末速度	m s⁻¹
a	Acceleration | 加速度	m s⁻²
t	Time | 时间	s

Remember that s is displacement, not distance. If an object changes direction, displacement can be zero or negative even though distance is positive.

请记住 s 是位移，而不是路程。如果物体改变运动方向，即使路程为正，位移也可能为零或为负。

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2. The Four Core Formulae | 四个核心公式

Edexcel expects you to recall and use these four equations. Each equation links four of the five SUVAT variables.

Edexcel 要求你熟记并运用以下四个方程。每个方程联系了五个 SUVAT 变量中的四个。

v = u + at

s = ((u + v) ÷ 2) × t

s = ut + ½ at²

v² = u² + 2as

A fifth equation, s = vt − ½ at², is also useful and comes from replacing u with v in the third formula. It is not always listed separately but can save time in some problems.

第五个公式 s = vt − ½ at² 也非常有用，它由第三个公式中将 u 替换为 v 得到。该公式并不总是单独列出，但在某些问题中可以节省时间。

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3. Deriving the Equations from Velocity-Time Graphs | 从速度-时间图推导公式

Under constant acceleration, a velocity-time graph is a straight line with gradient a. The first SUVAT equation is simply the gradient condition written algebraically.

在匀加速条件下，速度-时间图是一条斜率为 a 的直线。第一个 SUVAT 方程就是将斜率条件写成代数形式。

a = (v − u) ÷ t → v = u + at

The displacement is the area under the velocity-time graph. For constant acceleration, this area is a trapezium, giving the average velocity equation.

位移是速度-时间图下方的面积。对于匀加速运动，该面积是一个梯形，因此得到平均速度方程。

s = ½ (u + v) × t

Substituting v = u + at into the trapezium area formula gives s = ut + ½ at². Combining v = u + at with the average velocity formula gives v² = u² + 2as.

将 v = u + at 代入梯形面积公式可得到 s = ut + ½ at²。将 v = u + at 与平均速度公式联立可得到 v² = u² + 2as。

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4. Choosing the Right Equation | 选择正确的公式

To choose the correct equation, list the three quantities you know and the one quantity you need. Then select the equation that contains exactly those four variables.

要选择正确的公式，请列出你已知的三个量和需要求的一个量。然后选择恰好包含这四个变量的方程。

Equation | 方程	Missing variable | 不含的变量
v = u + at	s
s = ((u + v) ÷ 2) × t	a
s = ut + ½ at²	v
s = vt − ½ at²	u
v² = u² + 2as	t

A clear ‘known and unknown’ table is the fastest way to avoid choosing a formula with two unknown quantities.

清晰的“已知量与未知量”表格是避免选择含有两个未知量的公式的最快方法。

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5. Vertical Motion under Gravity | 重力作用下的竖直运动

When a particle moves vertically under gravity, the acceleration is due to gravity only, provided air resistance is ignored. On Earth, g = 9.8 m s⁻² downward.

当物体仅在重力作用下竖直运动时，若忽略空气阻力，加速度仅为重力加速度。在地球上，g = 9.8 m s⁻²，方向向下。

If the upward direction is taken as positive, then a = −9.8 m s⁻². At the highest point of a vertical throw, the velocity v is zero, but the acceleration is still −9.8 m s⁻².

如果取向上为正方向，则 a = −9.8 m s⁻²。在竖直上抛的最高点，速度 v 为零，但加速度仍为 −9.8 m s⁻²。

The displacement s in vertical motion represents height above the starting point. If the object falls below its starting point, s is negative.

竖直运动中的位移 s 表示物体相对于起点的高度。如果物体下落到起点以下，则 s 为负值。

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6. Sign Conventions and Direction | 符号约定与方向

SUVAT equations work in one dimension, so direction must be represented by signs. You may choose any positive direction, but you must use it consistently for u, v, a, and s.

SUVAT 方程适用于一维运动，因此方向必须用正负号表示。你可以选择任意正方向，但必须对 u、v、a 和 s 始终使用同一正方向。

• Choose a positive direction and keep it for the whole stage. | 选择正方向并在整个阶段中保持一致。
• Velocity is negative if it points opposite to the positive direction. | 如果速度指向正方向的反方向，则速度为负。
• Acceleration is negative if it acts opposite to the positive direction. | 如果加速度方向与正方向相反，则加速度为负。
• Speed is the magnitude of velocity, so speed is always non-negative. | 速率是速度的大小，因此速率始终为非负值。

In vertical motion, a common choice is upward positive. In horizontal motion, forward or right is usually positive, but this is not compulsory.

在竖直运动中，通常选择向上为正。在水平运动中，通常选择向前或向右为正，但这并不是强制性的。

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7. Worked Example: Horizontal Acceleration | 例题：水平加速

A car accelerates uniformly from 12 m s⁻¹ to 20 m s⁻¹ over a distance of 80 m. Find the acceleration and the time taken.

一辆汽车从 12 m s⁻¹ 匀加速到 20 m s⁻¹，经过距离 80 m。求加速度和所用时间。

Known quantities are u = 12 m s⁻¹, v = 20 m s⁻¹, and s = 80 m. The unknown is a, so use v² = u² + 2as.

已知量为 u = 12 m s⁻¹，v = 20 m s⁻¹，s = 80 m。未知量为 a，因此使用 v² = u² + 2as。

20² = 12² + 2 × a × 80

400 = 144 + 160a

a = 256 ÷ 160 = 1.6 m s⁻²

Now use v = u + at to find the time.

现在使用 v = u + at 求时间。

20 = 12 + 1.6 × t

t = 8 ÷ 1.6 = 5 s

The acceleration is 1.6 m s⁻² in the direction of motion, and the time taken is 5 s.

加速度为运动方向上的 1.6 m s⁻²，所用时间为 5 s。

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8. Worked Example: Vertical Projection | 例题：竖直上抛

A ball is thrown vertically upward with an initial speed of 14.7 m s⁻¹ from ground level. Find the maximum height, the time to reach it, and the total time in the air.

一个球以 14.7 m s⁻¹ 的初速度从地面竖直向上抛出。求最大高度、到达最大高度的时间以及球在空中运动的总时间。

Take upward as positive, so u = 14.7 m s⁻¹ and a = −9.8 m s⁻². At the maximum height, v = 0 m s⁻¹.

取向上为正方向，则 u = 14.7 m s⁻¹，a = −9.8 m s⁻²。在最大高度处，v = 0 m s⁻¹。

v² = u² + 2as → 0 = 14.7² + 2 × (−9.8) × s

0 = 216.09 − 19.6s → s = 11.025 m

Use v = u + at to find the time to maximum height.

使用 v = u + at 求到达最大高度的时间。

0 = 14.7 − 9.8t → t = 1.5 s

Since the ball returns to its starting height, the total time in the air is twice the time to maximum height, so total time = 3.0 s.

由于球返回到起始高度，在空中运动的总时间是到达最大高度时间的两倍，因此总时间为 3.0 s。

The returning velocity is v = 14.7 − 9.8 × 3.0 = −14.7 m s⁻¹, so the speed on return is 14.7 m s⁻¹.

返回时的速度为 v = 14.7 − 9.8 × 3.0 = −14.7 m s⁻¹，因此返回时的速率为 14.7 m s⁻¹。

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9. Two-Part Motion Problems | 两段运动问题

Many Edexcel exam questions involve two stages, such as a car accelerating and then braking, or a stone thrown upward and then falling down. You cannot apply one SUVAT equation across both stages if acceleration changes.

许多 Edexcel 考试题涉及两个阶段，例如汽车先加速后刹车，或石块先上升后下落。如果加速度发生变化，就不能在整个两段运动中直接使用一个 SUVAT 方程。

Split the motion into stages at the point where acceleration changes. The final velocity of the first stage often becomes the initial velocity of the second stage.

请在加速度改变的位置将运动分成多个阶段。第一阶段的末速度通常成为第二阶段的初速度。

• Stage 1: use SUVAT with its constant acceleration. | 阶段 1：在该阶段的恒定加速度下使用 SUVAT。
• Find the linking velocity or displacement. | 求出连接两个阶段的速度或位移。
• Stage 2: use SUVAT again with the new acceleration. | 阶段 2：用新的加速度再次使用 SUVAT。
• Add displacements and times carefully if required. | 如果需要，仔细相加位移和时间。

This staged approach is essential whenever acceleration is not constant over the whole journey.

只要整个运动过程中加速度不恒定，就必须采用这种分段处理的方法。

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10. Common Mistakes and Exam Tips | 常见错误与考试提示

Students often lose marks by using s as distance instead of displacement, or by applying the formulae when acceleration is not constant. In vertical motion, the most common error is giving g the wrong sign.

学生常常因为将 s 当作路程而不是位移，或在加速度不恒定时使用公式而失分。在竖直运动中，最常见的错误是给 g 添加了错误的符号。

• Always write down the three known quantities and the unknown first. | 始终先写出三个已知量和一个未知量。
• State the positive direction clearly in vertical motion. | 在竖直运动中清楚地说明正方向。
• At maximum height, v = 0, but a is still g. | 在最大高度处，v = 0，但 a 仍为 g。
• Check units: convert km h⁻¹ to m s⁻¹ when necessary. | 检查单位：必要时将 km h⁻¹ 转换为 m s⁻¹。
• If acceleration changes, split the motion into stages. | 如果加速度改变，将运动分段处理。

In an exam, a clear method and correct substitution earn most marks, even if the final arithmetic has a slip.

在考试中，清晰的方法和正确的代入过程可以赢得大部分分数，即使最终计算出现小错误也是如此。

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