📚 A-Level Physics: Kinematics Key Points | A-Level 物理：运动学考点精讲
Kinematics is the study of motion without considering its causes. It focuses on describing how objects move, using quantities like displacement, velocity, and acceleration. Mastering these concepts and the corresponding graphs and equations is fundamental to success in A‑Level Physics, as they form the basis for dynamics, projectile motion, and many other topics. This revision guide breaks down the essential kinematic ideas and problem‑solving techniques you need, with clear definitions and worked examples to help you tackle exam questions confidently.
运动学是研究物体运动而不涉及运动原因的学科，重点在于利用位移、速度和加速度等物理量来描述运动状态。掌握这些概念以及相应的图像和方程，是学好 A‑Level 物理的基础，因为它们为动力学、抛体运动等众多模块奠定根基。本篇复习指南梳理了必考的运动学核心概念与解题技巧，配合清晰的定义和例题，助你自信应对考试。
1. Scalars and Vectors | 标量与矢量
In kinematics, it is essential to distinguish between scalar and vector quantities. A scalar has magnitude only, such as distance, speed, and time. A vector has both magnitude and direction, such as displacement, velocity, and acceleration. Direction can be indicated by a positive or negative sign in one‑dimensional motion, or by an arrow or angle in two dimensions.
在运动学中，区分标量和矢量至关重要。标量只有大小，没有方向，例如路程、速率和时间。矢量既有大小又有方向，例如位移、速度和加速度。在一维运动中，方向可以用正负号表示；在二维运动中，方向常由箭头或角度表示。
When solving kinematic problems, always assign a positive direction (e.g. to the right or upwards) and treat any vector in the opposite direction as negative. This sign convention eliminates confusion when using equations of motion.
解题时务必先规定正方向（例如向右或向上为正），并把相反方向的矢量记为负值。这一正负号约定能避免在使用运动学方程时出现混淆。
2. Displacement, Velocity and Acceleration | 位移、速度与加速度
Displacement (s) is the straight‑line distance in a specified direction from the origin. It is a vector, measured in metres (m). Velocity (v) is the rate of change of displacement: v = Δs / Δt, given in m s⁻¹. Acceleration (a) is the rate of change of velocity: a = Δv / Δt, with units of m s⁻².
位移（s）是从起点指向终点的有向直线距离，是矢量，单位为米（m）。速度（v）是位移的变化率：v = Δs / Δt，单位是 m s⁻¹。加速度（a）是速度的变化率：a = Δv / Δt，单位是 m s⁻²。
Average velocity is total displacement divided by total time, whereas instantaneous velocity is the velocity at a specific instant (the gradient of a displacement‑time graph). Similarly, instantaneous acceleration is the gradient of a velocity‑time graph.
平均速度等于总位移除以总时间，而瞬时速度是某一时刻的速度（即位移‑时间图上的切线斜率）。类似地，瞬时加速度就是速度‑时间图上的斜率。
3. SUVAT Equations for Uniform Acceleration | 匀加速运动方程
When acceleration is constant, the five SUVAT equations link displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t). Each equation omits one variable, making it useful for different scenarios.
当加速度恒定时，五个 SUVAT 方程将位移（s）、初速度（u）、末速度（v）、加速度（a）和时间（t）联系起来。每个方程都缺少一个变量，适合不同的已知条件。
| Equation (omitted variable) | 方程（缺的变量） |
|---|---|
| v = u + at (s) | v = u + at（缺 s） |
| s = ut + ½ at² (v) | s = ut + ½ at²（缺 v） |
| v² = u² + 2as (t) | v² = u² + 2as（缺 t） |
| s = ½ (u + v) t (a) | s = ½ (u + v) t（缺 a） |
| s = vt – ½ at² (u) | s = vt – ½ at²（缺 u） |
To use these equations, you must identify three known quantities and the one you need, ensuring the motion is uniformly accelerated and the sign convention is consistent. Always convert units to SI first, and check that the answer is physically reasonable.
使用这些方程时，必须找出三个已知量和一个所求量，确认运动是匀加速且正负号一致。解题前先将单位统一为国际单位，并检查答案是否符合物理实际。
4. Free Fall and Vertical Motion | 自由落体与竖直运动
Free fall is the motion of an object under the influence of gravity alone, with negligible air resistance. The acceleration due to gravity is g = 9.81 m s⁻², directed downwards. In calculations, you may set upwards as positive, so a = –g = –9.81 m s⁻².
自由落体是指仅受重力作用、空气阻力可忽略的运动。重力加速度为 g = 9.81 m s⁻²，方向竖直向下。计算时可规定向上为正方向，那么 a = –g = –9.81 m s⁻²。
For an object thrown vertically upwards, its velocity decreases to zero at the highest point, after which it accelerates downwards. The time to rise equals the time to fall back to the launch height, and the whole motion can be analysed using SUVAT with a = –g.
对于竖直上抛的物体，速度在最高点减为零，随后向下加速。上升时间等于下落到抛出高度的时间，整个运动可用匀加速方程分析，只需令 a = –g。
5. Projectile Motion | 抛体运动
A projectile follows a parabolic path under constant downward acceleration g and zero horizontal acceleration (ignoring air resistance). The key is to resolve the initial velocity u into horizontal and vertical components: uₓ = u cos θ, uᵧ = u sin θ.
抛体在空气阻力可忽略的情况下，竖直方向受恒定加速度 g，水平方向加速度为零，其轨迹为抛物线。解题关键是先将初速度 u分解为水平分量 uₓ = u cos θ 和竖直分量 uᵧ = u sin θ。
Horizontally: constant speed, x = uₓ t. Vertically: uniform acceleration, y = uᵧ t – ½ g t² (taking up as positive). The time of flight is determined solely by the vertical motion. At the highest point, the vertical velocity is zero.
水平方向：匀速直线运动，x = uₓ t。竖直方向：匀加速运动，y = uᵧ t – ½ g t²（以向上为正）。飞行时间完全由竖直运动决定。在最高点处，竖直分速度为零。
6. Displacement–Time Graphs | 位移–时间图
A displacement–time graph directly shows how position changes with time. The gradient at any point gives the instantaneous velocity. A straight line means constant velocity; a curved line means changing velocity (acceleration). A horizontal line indicates the object is stationary.
位移–时间图直接展示了位置随时间的变化关系。图上任意点的斜率等于瞬时速度。直线表示匀速运动；曲线表示速度在变化（存在加速度）；水平线则表示物体静止。
The area under a displacement–time graph has no physical significance for velocity or acceleration; the key information comes from the gradient. If the graph becomes steeper, the speed is increasing; if it flattens, the speed is decreasing.
位移–时间图下方的面积对于速度或加速度没有物理意义；关键信息在于斜率。图线变陡，说明速率在增大；图线趋于平缓，说明速率在减小。
7. Velocity–Time Graphs | 速度–时间图
A velocity–time graph is one of the most powerful tools in kinematics. The gradient equals acceleration, and the area between the graph and the time axis represents displacement (areas below the axis count as negative displacement).
速度–时间图是运动学中最强大的工具之一。图线的斜率等于加速度，图线与时间轴围成的面积代表位移（时间轴以下的面积计为负位移）。
If the line slopes upward, the object is accelerating; a downward slope indicates deceleration (or acceleration in the negative direction). Zero slope means constant velocity. Counting squares or using geometric formulas can be used to find the area.
图线向上倾斜表示物体在加速；向下倾斜表示减速（或负方向的加速）。斜率为零表示匀速。求面积时可以采用数格子或几何公式的方法。
8. Acceleration–Time Graphs | 加速度–时间图
An acceleration–time graph plots acceleration against time. The area under the graph gives the change in velocity, Δv. For uniformly accelerated motion, the graph is a horizontal straight line; for variable acceleration, it will be a curve or sloped line.
加速度–时间图描绘了加速度随时间的变化。图线下的面积表示速度的变化量 Δv。匀加速运动对应的图线是一条水平直线；变加速运动则可能是曲线或斜线。
If the acceleration is zero, the velocity remains constant. A positive area adds speed, a negative area subtracts speed. This graph is particularly useful when acceleration is not constant and you need to find velocity change using integration or area estimation.
若加速度为零，则速度保持不变。正面积使速率增加，负面积使速率减小。当加速度不恒定时，利用该图通过积分或面积估算求速度变化显得尤为方便。
9. Variable Acceleration and Calculus | 变加速运动与微积分
When acceleration changes with time, you cannot use the SUVAT equations directly. Instead, calculus links displacement, velocity and acceleration: v = ds/dt, a = dv/dt = d²s/dt², and conversely s = ∫ v dt, v = ∫ a dt.
当加速度随时间变化时，不能直接套用匀加速方程。此时需要借助微积分来关联位移、速度和加速度：v = ds/dt，a = dv/dt = d²s/dt²，反之 s = ∫ v dt，v = ∫ a dt。
For example, if velocity is given by v = 3t² – 2t + 1, then acceleration is a = dv/dt = 6t – 2, and displacement can be found by integrating velocity with respect to time, including the initial conditions to find the constant of integration.
例如，若速度函数为 v = 3t² – 2t + 1，则加速度 a = dv/dt = 6t – 2，而位移可通过速度对时间积分得到，并利用初始条件确定积分常数。
Always remember to add the integration constant and evaluate it using boundary conditions, such as knowing the displacement at t=0. This approach is a key skill for solving problems with non‑uniform acceleration in A‑Level exams.
务必记得加上积分常数，并利用边界条件（如 t=0 时的位移）求出常数值。这是 A‑Level 考试中求解非匀加速问题的重要技能。
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