Deriving the Kinematic Equations of Motion | A-Level 物理：运动学方程的推导

📚 Deriving the Kinematic Equations of Motion | A-Level 物理：运动学方程的推导

The kinematic equations of motion describe the relationship between displacement, velocity, acceleration, and time for an object moving along a straight line with constant acceleration. These four equations form the foundation of many mechanics problems in A-Level Physics.

运动学方程描述了物体在直线上以恒定加速度运动时，位移、速度、加速度与时间之间的关系。这四个方程是 A-Level 物理中许多力学问题的基础。

1. Variables and Assumptions | 变量与假设

Before deriving the equations, we must define the symbols used throughout this article:

在推导方程之前，我们必须定义本文中使用的符号：

u – initial velocity (m s⁻¹)

u – 初速度（m s⁻¹）
v – final velocity (m s⁻¹)

v – 末速度（m s⁻¹）
a – constant acceleration (m s⁻²)

a – 恒定加速度（m s⁻²）
t – time interval (s)

t – 时间间隔（s）
s – displacement (m)

s – 位移（m）

These equations are valid only when the acceleration is constant. This means the net force acting on the object is constant, so the velocity changes at a uniform rate.

这些方程仅在加速度恒定有效。这意味着物体所受合力恒定，因此速度以均匀速率变化。

2. Deriving v = u + at from the Definition of Acceleration | 由加速度定义推导 v = u + at

Acceleration is defined as the rate of change of velocity. For constant acceleration, we can write:

加速度定义为速度的变化率。对于恒定加速度，我们可以写：

a = (v − u) / t

Multiply both sides by t:

两边同时乘以 t：

at = v − u

Then add u to both sides to obtain the first kinematic equation:

然后两边同时加上 u，得到第一个运动学方程：

v = u + at

This equation tells us how the final velocity depends on the initial velocity, the acceleration, and the time elapsed. It does not involve displacement.

该方程告诉我们末速度如何取决于初速度、加速度和经过的时间。它不涉及位移。

3. Displacement as the Area Under a Velocity-Time Graph | 位移作为速度-时间图像下的面积

For motion in one dimension, the displacement is equal to the area between the velocity-time graph and the time axis. This is true for any motion, whether acceleration is constant or not, because each small strip of width dt contributes v × dt to the displacement.

在一维运动中，位移等于速度-时间图像与时间轴之间的面积。这对于任何运动都成立，无论加速度是否恒定，因为每一小段宽度 dt 对应的面积贡献为 v × dt，即位移。

For constant acceleration, the velocity-time graph is a straight line segment. If the object starts with velocity u and ends with velocity v after time t, the graph forms a trapezoid.

对于恒定加速度，速度-时间图像是一条直线段。如果物体在时间 t 内从速度 u 变为速度 v，图像形成一个梯形。

The area of this trapezoid is the displacement:

该梯形的面积即为位移：

s = ½ × (u + v) × t

This is often written as:

这通常写作：

s = ½(u + v)t

This equation is particularly useful when the acceleration is not known but the initial and final velocities are given.

当加速度未知但初速度和末速度已知时，这个方程特别有用。

4. Deriving s = ut + ½at² from the Graph | 从图像推导 s = ut + ½at²

The area under the velocity-time graph can also be split into a rectangle and a triangle. The rectangle has height u and width t, representing the displacement that would occur if the velocity remained constant at u.

速度-时间图像下的面积也可以分成一个矩形和一个三角形。矩形的高为 u，宽为 t，代表如果速度保持 u 不变时所产生的位移。

The rectangle area is:

矩形的面积为：

Area of rectangle = u × t

The triangle has base t and height (v − u). Since v = u + at, the height is simply at. Therefore:

三角形的底为 t，高为 (v − u)。由于 v = u + at，所以高度就是 at。因此：

Area of triangle = ½ × t × at = ½at²

Adding the two areas gives the total displacement:

将两个面积相加得到总位移：

s = ut + ½at²

Alternatively, substitute v = u + at into s = ½(u + v)t:

或者，将 v = u + at 代入 s = ½(u + v)t：

s = ½(u + u + at)t = ut + ½at²

This equation is useful when the final velocity is not known.

当末速度未知时，这个方程非常有用。

5. Deriving v² = u² + 2as by Eliminating Time | 通过消去时间推导 v² = u² + 2as

Start with the first equation:

从第一个方程开始：

v = u + at

Rearrange to make t the subject:

重新排列，使 t 成为研究对象：

t = (v − u) / a

Now substitute this expression for t into s = ½(u + v)t:

现在将 t 的表达式代入 s = ½(u + v)t：

s = ½(u + v) × (v − u) / a

Use the difference of squares identity:

利用平方差公式：

(u + v)(v − u) = v² − u²

Therefore:

因此：

s = (v² − u²) / 2a

Multiply both sides by 2a:

两边同时乘以 2a：

2as = v² − u²

Finally, rearrange to obtain:

最后整理得到：

v² = u² + 2as

This equation is independent of time and is extremely useful when the time interval is not given.

该方程与时间无关，当时间间隔未知时非常有用。

6. Derivation Using Average Velocity | 利用平均速度推导

For constant acceleration, the average velocity over a time interval is the arithmetic mean of the initial and final velocities:

对于恒定加速度，一段时间内的平均速度是初速度和末速度的算术平均值：

v_avg = ½(u + v)

Displacement is average velocity multiplied by time:

位移等于平均速度乘以时间：

s = v_avg × t = ½(u + v)t

Substitute v = u + at into this expression:

将 v = u + at 代入该表达式：

s = ½(u + u + at)t = ut + ½at²

This method provides an alternative route to the same result, confirming the consistency of the kinematic equations.

这种方法提供了推导同一结果的另一条路径，也验证了运动学方程的一致性。

7. A Calculus Approach: Differentiation and Integration | 微积分方法：微分与积分

Although calculus is not always required for CIE A-Level Physics, it is valuable to understand how these equations arise from integration. Acceleration is the derivative of velocity with respect to time:

虽然 CIE A-Level 物理不一定要求微积分，但理解这些方程如何通过积分产生是有价值的。加速度是速度对时间的导数：

a = dv/dt

Rearrange and integrate both sides from t = 0 to t = t:

重新排列并对两边从 t = 0 到 t = t 积分：

∫ dv = ∫ a dt → v = u + at

Then, since velocity is the derivative of displacement with respect to time:

然后，由于速度是位移对时间的导数：

v = ds/dt

Integrate again with the initial condition s = 0 at t = 0:

再次积分，并利用初始条件 t = 0 时 s = 0：

s = ∫ (u + at) dt = ut + ½at²

To derive v² = u² + 2as, we can use the chain rule:

为了推导 v² = u² + 2as，我们可以使用链式法则：

a = dv/dt = (dv/ds) × (ds/dt) = v × dv/ds

So:

因此：

∫ a ds = ∫ v dv → as = ½v² − ½u²

This gives:

由此得到：

v² = u² + 2as

8. Choosing the Correct Equation | 选择正确的方程

Each kinematic equation omits one of the five variables. By identifying which variables are given and which are required, you can choose the most direct equation.

每个运动学方程都省略了五个变量中的一个。通过识别已知量和所求量，你可以选择最直接的方程。

Equation	Missing variable	Use when…
v = u + at	s	no displacement needed
s = ½(u + v)t	a	no acceleration given
s = ut + ½at²	v	no final velocity needed
v² = u² + 2as	t	no time given

The corresponding Chinese table for quick reference:

对应中文速查表：

方程	未出现变量	适用情况
v = u + at	s	不需要位移
s = ½(u + v)t	a	未给出加速度
s = ut + ½at²	v	不需要末速度
v² = u² + 2as	t	未给出时间

9. Sign Conventions and Common Pitfalls | 正负号约定与常见误区

Displacement, velocity, and acceleration are vectors. In one-dimensional problems, you must choose a positive direction and apply it consistently.

位移、速度和加速度都是矢量。在一维问题中，你必须选择一个正方向，并保持一致地应用它。

If an object is moving upward and upward is taken as positive, then the acceleration due to gravity is negative because gravity acts downward. At the highest point of a projectile-like vertical motion, the velocity is zero but the acceleration is still 9.8 m s⁻² downward.

如果物体向上运动且取向上为正，那么重力加速度为负，因为重力向下作用。在类似抛体的竖直运动中，最高点的速度为零，但加速度仍为 9.8 m s⁻² 向下。

A common mistake is to substitute only the magnitude of acceleration and ignore the sign. Always draw a diagram and label the positive direction before applying any equation.

一个常见错误是只代入加速度的大小而忽略符号。在应用任何方程之前，务必画出图像并标出正方向。

Another common pitfall is mixing units. For example, if velocity is given in km h⁻¹, it must be converted to m s⁻¹ before substitution into these equations.

另一个常见错误是混用单位。例如，如果速度以 km h⁻¹ 给出，必须转换为 m s⁻¹ 才能代入这些方程。

10. Worked Example | 例题分析

Example: A car accelerates uniformly from rest to 20 m s⁻¹ over a distance of 100 m. Find the acceleration and the time taken.

例题：一辆汽车从静止开始匀加速，经过 100 m 后速度达到 20 m s⁻¹。求加速度和所需时间。

Solution: First, identify the known variables: u = 0, v = 20 m s⁻¹, s = 100 m, a = ? t = ?

解答：首先确定已知量：u = 0，v = 20 m s⁻¹，s = 100 m，a = ？，t = ？

Use the equation that does not involve time:

使用不含时间的方程：

v² = u² + 2as

20² = 0² + 2 × a × 100

400 = 200a → a = 2 m s⁻²

Now use v = u + at to find t:

现在利用 v = u + at 求 t：

20 = 0 + 2t → t = 10 s

Alternatively, the time can be found from s = ½(u + v)t, giving the same result.

也可以利用 s = ½(u + v)t 求出时间，得到相同结果。

11. Summary of the Kinematic Equations | 运动学方程总结

The four kinematic equations are connected: each one is a rearrangement or combination of the others. Mastering their derivations helps you remember them and understand their limitations.

这四个运动学方程是相互联系的：每一个都是其它方程的变形或组合。熟练掌握它们的推导有助于记忆并理解其适用条件。

v = u + at

s = ½(u + v)t

s = ut + ½at²

v² = u² + 2as

These equations apply only for constant acceleration. When acceleration varies with time or position, they are no longer valid and calculus or graphical methods must be used instead.

这些方程仅适用于恒定加速度。当加速度随时间或位置变化时，这些方程不再适用，必须改用微积分或图像方法。

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Deriving the Kinematic Equations of Motion | A-Level 物理：运动学方程的推导