📚 Uniformly Accelerated Linear Motion Formulas (Part 2) | 匀加速直线运动公式（二）
This article continues our exploration of uniformly accelerated linear motion, focusing on the derived equations that link displacement, velocity, acceleration, and time. We will examine the velocity-displacement relation, average velocity methods, and practical applications such as free fall and stopping distances.
本文继续深入探讨匀加速直线运动，侧重于联系位移、速度、加速度与时间的导出公式。我们将研究速度-位移关系、平均速度方法，以及自由落体和停车距离等实际应用。
1. The Four Core Equations Revisited | 四个核心公式回顾
The standard equations for uniformly accelerated motion assume constant acceleration. They are valid for scalar motion along a straight line, with positive and negative signs indicating direction.
匀加速直线运动的标准公式假设加速度恒定。它们适用于沿直线运动的标量情形，正负号表示方向。
- v = u + at — links final velocity, initial velocity, acceleration, and time.
- s = ut + ½at² — displacement as a function of time.
- s = ½(u + v)t — displacement using average velocity.
- v² = u² + 2as — velocity-displacement relation.
- v = u + at — 联系末速度、初速度、加速度和时间。
- s = ut + ½at² — 位移随时间的变化关系。
- s = ½(u + v)t — 利用平均速度求位移。
- v² = u² + 2as — 速度-位移关系。
2. Deriving v² = u² + 2as | 推导 v² = u² + 2as
Eliminate time from v = u + at and s = ut + ½at² to obtain a relation independent of time.
从 v = u + at 和 s = ut + ½at² 中消去时间，可得与时间无关的关系式。
t = (v - u) / a, s = u[(v - u)/a] + ½a[(v - u)/a]²
Simplifying gives:
化简得到：
v² = u² + 2as
This equation is especially useful when time is not given or not required.
该公式在已知条件或所求量不涉及时间时尤为有用。
3. Average Velocity and Displacement | 平均速度与位移
For uniform acceleration, average velocity is the arithmetic mean of initial and final velocities.
在匀加速运动中，平均速度等于初速度与末速度的算术平均值。
vavg = (u + v) / 2
Then displacement in time t is:
则时间 t 内的位移为：
s = vavg × t = ½(u + v)t
This form is convenient when acceleration is unknown.
当加速度未知时，这种形式非常便捷。
4. From Graphical Perspective | 从图像角度理解
On a velocity-time graph, the slope gives acceleration and the area under the graph gives displacement.
在速度-时间图像中，斜率表示加速度，图像下方的面积表示位移。
- Slope = Δv/Δt = a
- Area of trapezoid = ½(u + v)t = s
- 斜率 = Δv/Δt = a
- 梯形面积 = ½(u + v)t = s
Displacement = Area under v–t graph
位移 = v–t 图像下的面积
5. Free Fall: g as a Special Case | 自由落体：g 作为一种特例
Free fall near the Earth's surface is uniformly accelerated motion with a constant acceleration g (approximately 9.81 m/s² downward).
在地球表面附近的自由落体是匀加速直线运动，加速度为常量 g（约为 9.81 m/s²，方向向下）。
Replacing a by g and taking downward as positive, the equations become:
将 a 替换为 g，并取向下为正方向，得：
v = u + gt, s = ut + ½gt², v² = u² + 2gs
For an object dropped from rest (u = 0), after falling distance s the speed is v = √2gs.
对于从静止释放的物体（u = 0），下落距离 s 后速度为 v = √2gs。
6. Vertical Projection: Upward Motion | 竖直上抛运动
When an object is thrown upward, it experiences acceleration g downward. Taking upward as positive, the acceleration is −g.
当物体竖直向上抛出时，它受到向下的加速度 g。取向上为正，加速度为 −g。
- Time to reach maximum height: t = u / g
- Maximum height: H = u² / (2g)
- Total flight time until return to start: T = 2u / g
- 到达最高点的时间：t = u / g
- 最大高度：H = u² / (2g)
- 回到出发点的总飞行时间：T = 2u / g
At maximum height, the instantaneous velocity is zero.
在最高点，瞬时速度为零。
7. Reaction Time and Stopping Distance | 反应时间与刹车距离
In driving problems, the total stopping distance consists of the distance travelled during the driver's reaction time (constant speed) plus the braking distance (uniform deceleration).
在驾驶问题中，总停车距离包含驾驶员反应时间内匀速行驶的距离，加上匀减速刹车的距离。
at speed u, reaction time is t_reaction, and braking deceleration is a (magnitude). Then:
设汽车速度为 u，反应时间为 t_reaction，刹车减速度大小为 a，则：
Total distance = u × t_reaction + u² / (2a)
总距离 = u × t_reaction + u² / (2a)
Doubling the speed more than doubles the total stopping distance because the braking part scales with u².
速度加倍会使总停车距离超过两倍，因为刹车部分与 u² 成正比。
8. Multi-Phase Motion | 多阶段运动
Some problems involve a particle that undergoes several uniform accelerations. The key is to treat each phase separately, using the final velocity of one phase as the initial velocity of the next.
有些问题涉及质点经历多个匀加速阶段。关键是分别处理每个阶段，将前一阶段的末速度作为后一阶段的初速度。
| Phase | Initial velocity | Acceleration | Time | Final velocity |
| 1 | u₁ | a₁ | t₁ | v₁ = u₁ + a₁t₁ |
| 2 | u₂ = v₁ | a₂ | t₂ | v₂ = v₁ + a₂t₂ |
The total displacement is the sum of displacements in each phase.
总位移等于各阶段位移之和。
9. Solving Problems: A Systematic Strategy | 解题：系统化策略
For any uniformly accelerated motion problem, follow these steps:
解决任何匀加速直线运动问题，可按以下步骤进行：
- Identify the given and required variables.
- Choose a positive direction (up/down or left/right).
- Select the equation that contains the target variable and excludes unknown ones.
- Substitute values with consistent units.
- Interpret the sign of the result.
- 确定已知量和待求量。
- 选择正方向（上/下或左/右）。
- 选取包含目标量且不含未知量的公式。
- 代入数值，注意单位一致。
- 解释结果正负号的意义。
10. Worked Example: Braking | 实例：刹车问题
Problem: A car travelling at 20 m/s brakes uniformly and stops after 50 m. Find the deceleration and the time taken.
题目：一辆汽车以 20 m/s 行驶，刹车后匀减速在 50 m 内停下。求减速度和所用时间。
Using v² = u² + 2as with v = 0, u = 20, s = 50:
利用 v² = u² + 2as，其中 v = 0，u = 20，s = 50：
0 = 20² + 2 × a × 50 → a = −4 m/s²
The negative sign indicates deceleration. To find time, use v = u + at:
负号表示减速。求时间，使用 v = u + at：
0 = 20 + (−4) × t → t = 5 s
11. Common Pitfalls | 常见误区
- Using equations across phases with different accelerations without resetting variables.
- Ignoring sign convention — consistently assign a positive direction.
- Applying s = ut + ½at² when acceleration is not constant.
- Forgetting that v² = u² + 2as gives only the magnitude of velocity, not direction.
- 在不同加速度的阶段之间直接套用公式，而不重置变量。
- 忽略正负号规范——必须一致地指定正方向。
- 在加速度不恒定时仍使用 s = ut + ½at²。
- 忘记 v² = u² + 2as 只给出速度大小，不能给出方向。
12. Summary | 总结
The four kinematic equations are interchangeable tools for solving uniformly accelerated motion problems. Mastery of their derivation, graphical meaning, and application to free fall and braking helps you handle both straightforward and multi-stage scenarios.
四个运动学公式是解决匀加速直线运动问题的通用工具。掌握它们的推导、图像意义，以及自由落体与刹车等应用，能帮助你处理简单及多阶段问题。
v = u + at, s = ut + ½at², s = ½(u + v)t, v² = u² + 2as
Practise identifying the quickest route to the unknown, and always check the physical reasonableness of your answer.
练习快速找到通向未知量的最简洁路径，并始终检查答案在物理上是否合理。
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