Uniformly Accelerated Rectilinear Motion: Essential Formulas (II) | 匀加速直线运动基本公式（二）

📚 Uniformly Accelerated Rectilinear Motion: Essential Formulas (II) | 匀加速直线运动基本公式（二）

The study of uniformly accelerated rectilinear motion (UARM) is one of the most fundamental topics in kinematics. In this second installment, we move beyond the basic velocity-time relationship and explore the displacement-time equation, the velocity-displacement equation, and their powerful applications in problem solving.

匀加速直线运动是运动学中最基础的内容之一。在第二部分的讲解中，我们将在速度-时间关系的基础上，进一步探讨位移-时间公式、速度-位移公式及其在解题中的强大应用。

1. Recap of the First Set of Formulas | 第一组公式回顾

Before introducing new equations, let us recall the three primary quantities: initial velocity \(v_0\), acceleration \(a\), and time \(t\). The fundamental velocity-time relation is \(v = v_0 + at\), which describes how velocity changes linearly with time under constant acceleration.

在引入新公式之前，我们先回顾三个基本物理量：初速度 \(v_0\)、加速度 \(a\) 和时间 \(t\)。最基本的描述匀加速运动速度变化规律的公式是 \(v = v_0 + at\)，它表明在恒定加速度下，速度随时间线性变化。

Additionally, the average velocity for uniformly accelerated motion is the arithmetic mean of the initial and final velocities: \(\bar{v} = \frac{v_0 + v}{2}\). This simple yet powerful relationship serves as the bridge between velocity and displacement.

此外，匀加速直线运动的平均速度等于初速度与末速度的算术平均值：\(\bar{v} = \frac{v_0 + v}{2}\)。这个简洁而有力的关系是在速度与位移之间搭建桥梁的关键。

2. The Displacement-Time Equation | 位移-时间公式

Since displacement equals average velocity multiplied by time, we can substitute the average velocity expression into \(s = \bar{v} t\) to obtain the displacement-time equation. After substitution, the equation becomes \(s = v_0 t + \frac{1}{2} a t^2\).

由于位移等于平均速度乘以时间，我们将 \(\bar{v} = (v_0 + v)/2\) 代入 \(s = \bar{v} t\)，即可得到位移-时间公式。代入后可得 \(s = v_0 t + \frac{1}{2} a t^2\)。

s = v₀t + ½at²

This equation is one of the most important in all of kinematics. It tells us that displacement is a quadratic function of time, which explains why the position-time graph of UARM is always a parabola, not a straight line.

该公式是整个运动学中最重要的公式之一。它告诉我们位移是时间的二次函数，这也解释了为什么匀加速直线运动的位移-时间图像总是一条抛物线，而非直线。

3. Special Case: Starting from Rest | 特例：从静止出发

When an object starts from rest, the initial velocity equals zero, and the displacement equation simplifies to a remarkably clean form: \(s = \frac{1}{2} a t^2\). This is one of the most frequently tested forms in examinations.

当物体从静止开始运动时，初速度为零，位移公式便会简化为非常整洁的形式：\(s = \frac{1}{2} a t^2\)。这是考试中最常考查的形式之一。

For example, a ball dropped from a height experiences free fall with \(a = g \approx 9.8 \text{ m/s}^2\). After 2 seconds, the distance fallen is \(s = \frac{1}{2} \times 9.8 \times 2^2 = 19.6 \text{ m}\).

例如，一个从高处自由下落的球，其加速度 \(a = g \approx 9.8 \text{ m/s}^2\)。经过2秒后，下落距离为 \(s = \frac{1}{2} \times 9.8 \times 2^2 = 19.6 \text{ m}\)。

Students often confuse the distance fallen in the first second, second second, and third second. In free fall:

Distance in 1st second: \(s_1 = \frac{1}{2} g \times 1^2 = 4.9 \text{ m}\)
Distance in 2nd second: \(s_2 = \frac{1}{2} g \times 4 – \frac{1}{2} g \times 1 = 14.7 \text{ m}\)
Distance in 3rd second: \(s_3 = \frac{1}{2} g \times 9 – \frac{1}{2} g \times 4 = 24.5 \text{ m}\)

同学们常常混淆第1秒内、第2秒内和第3秒内下落的距离。在自由落体中：

第1秒内下落：\(s_1 = \frac{1}{2} g \times 1^2 = 4.9 \text{ m}\)
第2秒内下落：\(s_2 = \frac{1}{2} g \times 4 – \frac{1}{2} g \times 1 = 14.7 \text{ m}\)
第3秒内下落：\(s_3 = \frac{1}{2} g \times 9 – \frac{1}{2} g \times 4 = 24.5 \text{ m}\)

4. The Velocity-Displacement Equation | 速度-位移公式

Sometimes we are given the displacement and velocity rather than time. In such cases, eliminating time from the first two equations yields the velocity-displacement equation: \(v^2 = v_0^2 + 2as\).

有时我们已知位移和速度，而不涉及时间。在这种情况下，从前两个公式中消去时间，便可以得到速度-位移公式：\(v^2 = v_0^2 + 2as\)。

v² = v₀² + 2as

This equation is invaluable because it links velocity and displacement without requiring the time parameter. It is especially useful when solving problems where time is neither given nor requested, such as determining the stopping distance of a vehicle.

这个公式的价值在于它直接将速度与位移联系起来，完全不需要时间这个参数。当题目中既没有给出时间、也不要求计算时间时（例如求汽车的制动距离），该公式尤为实用。

Worked Example: A car travelling at 30 m/s applies brakes and decelerates uniformly at 5 m/s². What distance does it travel before stopping?

\(v^2 = v_0^2 + 2as \Rightarrow 0 = 30^2 + 2(-5)s \Rightarrow 0 = 900 – 10s \Rightarrow s = 90 \text{ m}\)

例题：一辆汽车以30 m/s的速度行驶，刹车后以5 m/s²的加速度匀减速。它停止前行驶了多远的距离？

\(v^2 = v_0^2 + 2as \Rightarrow 0 = 30^2 + 2(-5)s \Rightarrow 0 = 900 – 10s \Rightarrow s = 90 \text{ m}\)

5. Deriving the Formulas | 公式的推导

Understanding the derivation of these formulas is essential for mastering physics, as it reinforces the logical structure underlying all kinematics. Let us derive the three main equations systematically.

理解这些公式的推导过程对于掌握物理至关重要，因为这能强化运动学背后的逻辑结构。下面我们系统地推导三个主要公式。

Derivation from velocity-time graph. For UARM, the v-t graph is a straight line. The area under this line represents displacement. This area can be split into a rectangle and a triangle:

Rectangle area: \(v_0 \times t\)
Triangle area: \(\frac{1}{2} \times t \times (at) = \frac{1}{2} a t^2\)
Total: \(s = v_0 t + \frac{1}{2} a t^2\)

从速度-时间图像推导。在匀加速直线运动中，v-t 图像是一条直线，图线与时间轴围成的面积代表位移。这个面积可以拆分为一个矩形和一个三角形：

矩形面积：\(v_0 \times t\)
三角形面积：\(\frac{1}{2} \times t \times (at) = \frac{1}{2} a t^2\)
总面积：\(s = v_0 t + \frac{1}{2} a t^2\)

Deriving \(v^2 = v_0^2 + 2as\). From \(v = v_0 + at\), we solve for \(t = (v – v_0)/a\). Substituting this into \(s = v_0 t + \frac{1}{2} a t^2\) and simplifying algebraically yields \(v^2 = v_0^2 + 2as\).

推导 \(v^2 = v_0^2 + 2as\)。由 \(v = v_0 + at\)，可解出 \(t = (v – v_0)/a\)。将其代入 \(s = v_0 t + \frac{1}{2} a t^2\) 并进行代数化简，即可得到 \(v^2 = v_0^2 + 2as\)。

6. Choosing the Right Formula | 如何选择合适的公式

One of the most common difficulties students encounter is selecting the correct equation from the set of five UARM formulas. The key is to identify which variables are given and which variable you need to find.

同学们最常遇到的困难之一，是从五个匀加速直线运动公式中选出正确的那个。关键在于判断哪些变量是已知的，以及你要求的是哪个变量。

The five standard symbols are: \(s\), \(u\) (or \(v_0\)), \(v\), \(a\), and \(t\). Each equation omits exactly one of these variables:

五个标准物理量为：\(s\)、\(u\)（或 \(v_0\)）、\(v\)、\(a\) 和 \(t\)。每个公式恰好省略了其中一个变量：

Equation 公式	Omits 省略	Best used when 最佳使用场景
v = u + at	s	No displacement involved 不涉及位移
s = ½(u + v)t	a	No acceleration involved 不涉及加速度
s = ut + ½at²	v	No final velocity involved 不涉及末速度
v² = u² + 2as	t	No time involved 不涉及时间

To select the appropriate formula, list the known variables, identify the unknown, then choose the equation that contains exactly the variables you have and the one you need.

选择公式时，先列出已知量，明确未知量，然后选择恰好包含所有已知量和所求量的那个公式。

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

In UARM problems, the direction of motion matters enormously. A common convention is to take the initial direction of motion as positive. Quantities acting in the opposite direction — such as acceleration during braking, or displacement when an object returns to its starting point — must be assigned negative values.

在匀加速直线运动问题中，运动方向至关重要。通常约定取初速度方向为正方向。与此方向相反的物理量——例如刹车时的加速度，或物体回到出发点时的位移——必须赋予负值。

Example: A ball is thrown vertically upward with an initial velocity of 20 m/s. Taking upward as positive, the acceleration is \(a = -9.8 \text{ m/s}^2\). After 5 seconds, the displacement is \(s = 20 \times 5 + \frac{1}{2}(-9.8)(5^2) = 100 – 122.5 = -22.5 \text{ m}\).

例题：小球以20 m/s的初速度竖直上抛。取向上为正方向，则加速度为 \(a = -9.8 \text{ m/s}^2\)。经过5秒后，位移为 \(s = 20 \times 5 + \frac{1}{2}(-9.8)(5^2) = 100 – 122.5 = -22.5 \text{ m}\)。

The negative sign indicates that the ball is 22.5 m below its starting point. Failing to apply proper sign conventions is one of the most frequent sources of error in physics examinations.

负号表示小球在出发点下方22.5米处。未能正确应用符号约定是物理考试中最常见的失分原因之一。

8. The Odd-Ratio Rule: Equal Time Intervals | 奇数比规律：等时间间隔

For an object starting from rest, there is a fascinating pattern in the distances travelled in successive equal time intervals. The displacement ratios follow 1 : 3 : 5 : 7 : … , the sequence of odd numbers.

对于从静止开始运动的物体，在连续相等的时间间隔内行进的距离存在一个有趣的规律：位移之比满足 1 : 3 : 5 : 7 : …，即奇数序列。

Proof: The displacement in the first interval is \(\frac{1}{2}a(1)^2 = \frac{1}{2}a\). The displacement in the first two intervals is \(\frac{1}{2}a(2)^2 = 2a\), so the displacement in the second interval alone is \(2a – \frac{1}{2}a = \frac{3}{2}a\). Continuing this logic, the displacements in the first three intervals are \(a/2, 3a/2, 5a/2\), giving the ratio 1 : 3 : 5.

证明：第1个时间间隔内的位移为 \(\frac{1}{2}a(1)^2 = \frac{1}{2}a\)。前2个时间间隔的总位移为 \(\frac{1}{2}a(2)^2 = 2a\)，因此第2个间隔内的位移为 \(2a – \frac{1}{2}a = \frac{3}{2}a\)。依此类推，前3个间隔内的位移分别为 \(a/2、3a/2、5a/2\)，之比为 1 : 3 : 5。

This rule provides a quick shortcut in many problems. For instance, if a falling object travels 5 m in the first second, it will travel 15 m in the second second, 25 m in the third second, and so on — without any additional calculation.

这个规律为许多问题提供了快速解法。例如，如果一个自由落体在第1秒内下落5米，那么第2秒内将下落15米，第3秒内下落25米，以此类推——无需再进行额外计算。

9. Equal Displacement Rule: Time in Ratio | 等位移时间比规律

Just as there is a pattern for equal time intervals, there is also a pattern for equal displacement intervals. For an object starting from rest, the times taken to traverse successive equal displacements follow the ratio 1 : (√2 – 1) : (√3 – √2) : (√4 – √3) : …

与等时间间隔的规律类似，等位移间隔也存在规律。对于从静止开始运动的物体，通过连续相等位移所用的时间之比为 1 : (√2 – 1) : (√3 – √2) : (√4 – √3) : …

Derivation: The time to travel the first displacement \(s\) is found from \(s = \frac{1}{2} a t_1^2\), giving \(t_1 = \sqrt{2s/a}\). The time to travel the first \(2s\) is \(t_2 = \sqrt{4s/a} = \sqrt{2} \cdot t_1\). The time for the second segment alone is \(t_2 – t_1 = (\sqrt{2} – 1)t_1\), and so on.

推导：通过第一段位移 \(s\) 所需时间可由 \(s = \frac{1}{2} a t_1^2\) 求出，即 \(t_1 = \sqrt{2s/a}\)。通过前 \(2s\) 位移所需时间为 \(t_2 = \sqrt{4s/a} = \sqrt{2} \cdot t_1\)。通过第二段位移单独所需时间为 \(t_2 – t_1 = (\sqrt{2} – 1)t_1\)，依此类推。

This ratio rule is particularly useful in free-fall problems where an object falls through equal vertical distances, such as a ball passing successive windows on a building.

这个比例规律在自由落体问题中尤为有用，例如小球依次经过大楼各扇窗户的情形。

10. Graphical Interpretation | 图像法解读

The three UARM formulas have direct graphical interpretations that deepen our understanding. The velocity-time graph is a straight line whose slope equals the acceleration \(a\), and whose area under the curve equals displacement \(s\).

三个匀加速直线运动公式都有直接的图像意义，能加深我们的理解。速度-时间图像是一条直线，其斜率等于加速度 \(a\)，图线下方的面积等于位移 \(s\)。

In the acceleration-time graph for UARM, the graph is a horizontal straight line at \(a\), and the area under this line equals the change in velocity. Meanwhile, the displacement-time graph is a parabola whose vertex and orientation depend on the sign of acceleration and the initial velocity.

匀加速直线运动的加速度-时间图像是一条高度为 \(a\) 的水平直线，图线下方围成的面积等于速度的变化量。同时，位移-时间图像是一条抛物线，其顶点位置和开口方向取决于加速度的符号和初速度的大小。

Students who master these graphical relationships can often solve problems faster and catch conceptual errors that pure algebra would miss. For example, a displacement-time graph that is not a parabola immediately signals that the motion is not uniformly accelerated.

掌握了这些图像关系的同学往往能更快速地解题，并且能发现纯代数计算中容易忽略的概念性错误。例如，如果位移-时间图像不是抛物线，则立即表明该运动不是匀加速直线运动。

11. Common Mistakes and Pitfalls | 常见错误与易错点

Several errors appear repeatedly in student solutions to UARM problems. Being aware of them in advance is the first step toward avoiding them in examinations.

在匀加速直线运动问题中，有若干错误在学生的解答中反复出现。提前了解这些错误，是考场上避免它们的第一步。

Mistake 1: Confusing distance and displacement. Distance is the total length of the path travelled, always positive, while displacement is a vector quantity that can be negative. In problems involving direction reversal, these two quantities differ significantly.

错误一：混淆路程与位移。路程是路径的总长度，永远为正；位移是矢量，可以为负。在涉及方向改变的问题中，两者相差巨大。

Mistake 2: Using UARM formulas for non-uniform acceleration. These formulas only apply when acceleration is constant. If acceleration changes, the equations are no longer valid.

错误二：在非匀加速运动中使用匀加速公式。这些公式仅在加速度恒定时成立。如果加速度发生变化，公式将不再适用。

Mistake 3: Incorrect sign of acceleration. During braking, acceleration is negative relative to the direction of motion. Writing \(a = +5 \text{ m/s}^2\) for a decelerating car can lead to absurd results such as negative stopping distances.

错误三：加速度符号错误。在刹车过程中，加速度相对于运动方向为负。如果将减速汽车的加速度写成 \(a = +5 \text{ m/s}^2\)，可能得出负的制动距离等荒谬结果。

12. Practice Problems | 练习与巩固

The best way to internalize these formulas is through practice. Below are two classic problems that test the application of the formulas covered in this article.

深化理解这些公式的最佳途径是练习。下面是两道经典题目，用于检验本篇文章所讲公式的灵活运用。

Question 1: A particle starts from rest and accelerates uniformly at 2 m/s² for 6 seconds. It then continues at constant velocity for 4 seconds. Find the total displacement.

题目一：一质点从静止出发，以2 m/s²的加速度匀加速运动6秒，然后以匀速继续运动4秒。求总位移。

Solution: Phase 1: \(s_1 = \frac{1}{2} \times 2 \times 6^2 = 36 \text{ m}\), \(v = 2 \times 6 = 12 \text{ m/s}\). Phase 2: \(s_2 = 12 \times 4 = 48 \text{ m}\). Total: \(s = 36 + 48 = 84 \text{ m}\).

解答：第一阶段：\(s_1 = \frac{1}{2} \times 2 \times 6^2 = 36 \text{ m}\)，\(v = 2 \times 6 = 12 \text{ m/s}\)。第二阶段：\(s_2 = 12 \times 4 = 48 \text{ m}\)。总位移：\(s = 36 + 48 = 84 \text{ m}\)。

Question 2: A stone is thrown vertically upward with a velocity of 24.5 m/s. Using \(g = 9.8 \text{ m/s}^2\), calculate the maximum height reached.

题目二：一块石头以24.5 m/s的初速度竖直上抛。取 \(g = 9.8 \text{ m/s}^2\)，求上升的最大高度。

Solution: At maximum height, \(v = 0\). Using \(v^2 = u^2 + 2as\): \(0 = 24.5^2 + 2(-9.8)s \Rightarrow 0 = 600.25 – 19.6s \Rightarrow s = 30.625 \text{ m}\).

解答：在最高点处，\(v = 0\)。使用 \(v^2 = u^2 + 2as\)：\(0 = 24.5^2 + 2(-9.8)s \Rightarrow 0 = 600.25 – 19.6s \Rightarrow s = 30.625 \text{ m}\)。

These problems demonstrate how the choice of formula depends critically on which variables are provided. In Question 1, time is central and displacement is found through the time-based equation; in Question 2, time is not involved at all, making the velocity-displacement equation the natural choice.

这两道题展示了公式选择如何取决于题给条件。题目一以时间为主线，通过含时间的公式求位移；题目二完全不涉及时间，因此速度-位移方程是自然而然的选择。

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Uniformly Accelerated Rectilinear Motion: Essential Formulas (II) | 匀加速直线运动基本公式（二）