The Equations of Motion | 运动方程

📚 The Equations of Motion | 运动方程

In CIE A-Level Physics, the equations of motion describe how displacement, initial velocity, final velocity, acceleration, and time are related when acceleration is constant. They are often called the ‘suvat’ equations because they link s, u, v, a, and t.

在 CIE A-Level 物理中，运动方程描述在加速度恒定的条件下，位移、初速度、末速度、加速度和时间之间的关系。由于它们联系 s、u、v、a、t 五个量，常被称为 “suvat” 方程。

1. Kinematic Quantities and Sign Conventions | 运动学物理量与正方向约定

The five key quantities are displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t). Displacement, velocity and acceleration are vectors, so their signs depend on the chosen positive direction.

五个关键物理量是位移 s、初速度 u、末速度 v、加速度 a 和时间 t。位移、速度和加速度都是矢量，因此它们的正负取决于所选的正方向。

In one-dimensional motion, it is essential to choose a clear positive direction before substituting values. A quantity is positive if it points along that direction and negative if it points opposite.

在一维运动中，代入数值前必须先选定正方向。若某矢量的方向与正方向相同则为正，相反则为负。

The equations of motion are valid only when acceleration is constant. For changing acceleration, you should use calculus or graph methods.

运动方程仅在加速度恒定时成立。若加速度变化，应使用微积分或图像方法。

2. The Four SUVAT Equations | 四个 SUVAT 方程

There are four standard equations. Each equation omits one of the five quantities, so you can choose the equation that does not involve the quantity you are not given or not asked for.

四个标准方程各自少一个物理量，因此可以选择不含有题目未给或未求物理量的方程。

The four equations are shown below.

四个方程如下所示。

v = u + at

s = ½(u + v)t

s = ut + ½at²

v² = u² + 2as

The table below shows which quantity each equation omits.

下表显示每个方程省略的物理量。

Equation	Missing quantity	Typical use
v = u + at	s	finding final velocity with no displacement
s = ½(u + v)t	a	using average velocity when acceleration is not given
s = ut + ½at²	v	finding displacement without final velocity
v² = u² + 2as	t	relating velocity and displacement without time

3. Deriving the Equations from Definitions | 从定义推导运动方程

The first equation follows directly from the definition of constant acceleration: a = (v – u) / t, which rearranges to v = u + at.

第一个方程由匀加速度定义 a = (v – u) / t 直接得出，整理得 v = u + at。

The average velocity during uniform acceleration is ½(u + v). Multiplying this average velocity by time gives displacement, so s = ½(u + v)t.

匀加速过程中的平均速度为 ½(u + v)。用平均速度乘以时间得到位移，因此 s = ½(u + v)t。

Substituting v = u + at into s = ½(u + v)t gives s = ut + ½at². Substituting t = (v – u)/a into s = ½(u + v)t gives v² = u² + 2as.

将 v = u + at 代入 s = ½(u + v)t 得到 s = ut + ½at²。将 t = (v – u)/a 代入 s = ½(u + v)t 得到 v² = u² + 2as。

4. Choosing the Positive Direction in Problems | 解题时选择正方向

Before solving a problem, draw a simple diagram and mark the positive direction with an arrow. All vector quantities such as u, v, a and g must be given signs according to this arrow.

解题前先画简图，用箭头标出正方向。所有矢量如 u、v、a 和 g 都必须按照该箭头取正负号。

For example, if upward is chosen as positive, then the acceleration due to gravity is g = -9.81 m s⁻². If downward is positive, then g = +9.81 m s⁻².

例如，若取向上为正，则重力加速度 g = -9.81 m s⁻²；若取向下为正，则 g = +9.81 m s⁻²。

Do not change the positive direction halfway through a problem. Keeping one convention prevents sign errors in multistep calculations.

解题过程中不要中途改变正方向。保持同一个约定可以避免多步计算中的符号错误。

5. Worked Example: Uniform Acceleration | 例题：匀加速直线运动

A car accelerates uniformly from rest at 3.0 m s⁻² for 8.0 s. Calculate the final velocity and the displacement.

一辆汽车从静止开始以 3.0 m s⁻² 的加速度匀加速运动 8.0 s。求末速度和位移。

Choose the direction of motion as positive. Given u = 0, a = 3.0 m s⁻², t = 8.0 s. Use v = u + at: v = 0 + (3.0)(8.0) = 24 m s⁻¹.

取运动方向为正。已知 u = 0、a = 3.0 m s⁻²、t = 8.0 s。使用 v = u + at：v = 0 + (3.0)(8.0) = 24 m s⁻¹。

Then use s = ut + ½at²: s = (0)(8.0) + ½(3.0)(8.0)² = 96 m.

再用 s = ut + ½at²：s = (0)(8.0) + ½(3.0)(8.0)² = 96 m。

You could also use s = ½(u + v)t because the average velocity is 12 m s⁻¹. This gives s = 12 × 8.0 = 96 m, confirming the answer.

也可用 s = ½(u + v)t，因为平均速度为 12 m s⁻¹。由此 s = 12 × 8.0 = 96 m，验证答案。

6. Free Fall Under Gravity | 重力作用下的自由落体

A ball is thrown vertically upward with speed 15 m s⁻¹ from a point 2.0 m above the ground. Take upward as positive and g = 9.81 m s⁻². Find the maximum height above the ground.

一个球以 15 m s⁻¹ 的初速度从离地 2.0 m 处竖直上抛。取向上为正，g = 9.81 m s⁻²。求球离地面的最大高度。

At maximum height the final velocity is zero. Use v² = u² + 2as with u = 15 m s⁻¹, v = 0, a = -9.81 m s⁻².

在最高点末速度为零。使用 v² = u² + 2as，其中 u = 15 m s⁻¹、v = 0、a = -9.81 m s⁻²。

0 = (15)² + 2(-9.81)s, so s = 225 / (2 × 9.81) = 11.5 m above the launch point. The height above the ground is therefore 11.5 + 2.0 = 13.5 m.

0 = (15)² + 2(-9.81)s，因此 s = 225 / (2 × 9.81) = 11.5 m（抛出点上方）。离地高度为 11.5 + 2.0 = 13.5 m。

7. Projectile Motion: Two-Dimensional Motion | 抛体运动：二维运动

In projectile motion, the horizontal and vertical motions are independent. The equations of motion are applied separately to the vertical direction, while horizontal velocity is constant if air resistance is negligible.

在抛体运动中，水平与竖直运动相互独立。运动方程分别应用于竖直方向；若忽略空气阻力，水平速度保持不变。

For a projectile launched horizontally from height h, the time to reach the ground is found from h = ½gt² in the vertical direction. The horizontal range is then x = u t, where u is the constant horizontal speed.

对于从高度 h 水平抛出的物体，先由竖直方向 h = ½gt² 求出落地时间，再由水平方向 x = u t 求出水平射程，其中 u 是恒定的水平速度。

For a projectile at angle θ to the horizontal, resolve the initial velocity into u cos θ horizontally and u sin θ vertically. Use the vertical equations of motion to find time of flight, maximum height, or final vertical velocity.

对于与水平成 θ 角抛出的物体，将初速度分解为水平分量 u cos θ 和竖直分量 u sin θ。使用竖直运动方程求飞行时间、最大高度或末竖直分速度。

8. Motion Graphs and the Equations of Motion | 运动图像与运动方程

The equations of motion are closely linked to velocity-time graphs. For constant acceleration, the velocity-time graph is a straight line with gradient a and area under the graph equal to displacement s.

运动方程与速度-时间图像密切相关。匀加速运动的速度-时间图像是一条直线，其斜率等于加速度 a，图像下面积等于位移 s。

The equation v = u + at simply expresses the gradient: final velocity is the initial velocity plus

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The Equations of Motion | 运动方程