📚 A-Level AQA Maths: Kinematics Essentials | AQA A-Level 数学运动学考点精讲
Kinematics is the branch of mechanics concerned with describing motion without considering the forces that cause it. In AQA A-Level Mathematics, a solid understanding of straight-line motion, SUVAT equations, motion graphs, calculus-based methods, and projectile motion is essential for success in the applied paper.
运动学是力学的一个分支，专注于描述运动而不考虑引起运动的力。在AQA A-Level数学中，你必须熟练掌握直线运动、SUVAT方程、运动图线、基于微积分的方法以及抛体运动，才能在应用卷中取得好成绩。
1. Displacement, Velocity and Acceleration | 位移、速度与加速度
Displacement (s) is a vector quantity that measures the change in position relative to a fixed origin, with both magnitude and direction. Distance, however, is a scalar and only measures the path length travelled.
位移 (s) 是一个矢量，它测量相对于固定原点的位置变化，具有大小和方向。而路程是一个标量，只测量经过的路径长度。
Velocity (v) is the rate of change of displacement with respect to time, also a vector. Acceleration (a) is the rate of change of velocity. On a displacement-time graph, velocity is the gradient; on a velocity-time graph, acceleration is the gradient and displacement is the area under the curve.
速度 (v) 是位移随时间的变化率，也是矢量。加速度 (a) 是速度的变化率。在位移-时间图上，速度等于斜率；在速度-时间图上，加速度等于斜率，位移等于曲线下的面积。
Standard units used in mechanics are metres (m) for displacement, metres per second (m s⁻¹) for velocity, and metres per second squared (m s⁻²) for acceleration. When using vector equations, a positive direction must be defined.
力学中使用的标准单位是：位移用米 (m)，速度用米每秒 (m s⁻¹)，加速度用米每二次方秒 (m s⁻²)。使用矢量方程时，必须规定正方向。
2. The Five SUVAT Equations | 五个 SUVAT 方程
For motion in a straight line with constant acceleration, five equations link the variables s (displacement), u (initial velocity), v (final velocity), a (acceleration), and t (time). Each equation omits one variable, enabling you to choose the correct formula based on the known quantities.
对于匀加速直线运动，有五个方程将变量 s（位移）、u（初速度）、v（末速度）、a（加速度）和 t（时间）联系起来。每个方程省略一个变量，因此你可以根据已知量选择正确的公式。
| Equation | Missing variable |
| v = u + at | s |
| s = ut + ½ at² | v |
| s = ½ (u + v) t | a |
| v² = u² + 2as | t |
| s = vt − ½ at² | u |
You must memorise these equations and recognise which variable is not involved in order to apply the correct one efficiently. Always substitute known values carefully, including signs for direction.
你必须熟记这些方程，并能识别哪个变量未涉及，以便高效地选用正确的方程。代入已知量时务必仔细，包括表明方向的符号。
These equations apply only when the acceleration is constant. If the acceleration varies, you must use calculus techniques instead.
这些方程仅在加速度恒定时适用。如果加速度变化，必须改用微积分方法。
3. Choosing a Positive Direction and Sign Conventions | 选择正方向与符号约定
When modelling motion, you must define a positive direction (e.g. upwards or to the right). All vector quantities must then be assigned signs accordingly. For example, if upward is positive, gravitational acceleration is −g, where g = 9.8 m s⁻².
在建立运动模型时，你必须定义一个正方向（例如向上或向右）。然后所有矢量都要相应地赋予符号。例如，如果向上为正，重力加速度就是 −g，其中 g = 9.8 m s⁻²。
A common mistake is to treat all velocities and accelerations as positive. A deceleration is simply an acceleration with a sign opposite to the velocity. Use consistent signs throughout a calculation.
一个常见错误是把所有速度和加速度都当作正值。减速其实就是加速度符号与速度相反。整个计算过程必须保持符号一致。
For vertical motion under gravity, choosing upward as positive gives a constant acceleration a = −9.8 m s⁻². An object thrown upwards will have an initial positive velocity, slowing down until v = 0 at the maximum height.
对于重力作用下的竖直运动，选取向上为正，则恒定加速度 a = −9.8 m s⁻²。向上抛出的物体初速度为正，速度逐渐减小，直到最高点 v = 0。
4. Vertical Motion Under Gravity | 重力作用下的竖直运动
Problems involving an object falling or being projected vertically can be solved with SUVAT, using a = −g (if upward is positive) or a = +g (if downward is positive). The value of g is taken as 9.8 m s⁻² unless specified otherwise.
涉及物体竖直下落或竖直上抛的问题可用 SUVAT 方程求解，取 a = −g（若向上为正）或 a = +g（若向下为正）。除非题目另有说明，g 的值取 9.8 m s⁻²。
Typical questions ask for maximum height reached, time of flight, speed on hitting the ground, or the total displacement. At the maximum height the instantaneous velocity is zero, a crucial piece of information.
典型的题目会求最大高度、飞行时间、落地速度或总位移。在最高点，瞬时速度为零，这是一个关键信息。
For a particle thrown upwards from ground level with speed u, the maximum height is h = u² / (2g) and the time to reach it is t = u / g. The total time of flight until returning to the ground is 2u / g.
对于从地面以初速 u 竖直向上抛出的物体，最大高度为 h = u² / (2g)，到达最高点的时间为 t = u / g。落回地面的总飞行时间为 2u / g。
5. Displacement-Time and Velocity-Time Graphs | 位移-时间图与速度-时间图
A displacement-time graph has time on the horizontal axis and displacement on the vertical axis. The gradient gives the velocity. A straight line means constant velocity; a curve indicates changing velocity, and a horizontal segment means the object is stationary.
位移-时间图以时间为横轴、位移为纵轴。斜率表示速度。直线表示匀速；曲线表示速度在变化；水平线段表示物体静止。
A velocity-time graph provides even more information: the gradient gives the acceleration, and the area between the graph and the time axis gives the displacement. Negative velocity indicates motion in the opposite direction.
速度-时间图提供更多信息：斜率表示加速度，图线与时间轴之间的面积表示位移。速度为负表示向反方向运动。
You should be able to sketch and interpret these graphs, converting between displacement, velocity, and acceleration descriptions. For constant acceleration, the velocity-time graph is a straight line, and the area under it is easily calculated as trapezium.
你应该能够绘制和解释这些图，并在位移、速度和加速度描述之间转换。对于匀加速运动，速度-时间图是一条直线，其下的面积可用梯形公式轻松计算。
6. Introducing Calculus into Kinematics | 引入微积分处理运动学
When acceleration is not constant, you must use differentiation and integration to connect displacement, velocity, and acceleration. The fundamental relationships are:
当加速度不恒定时，你必须使用微分和积分来联系位移、速度与加速度。基本关系为：
v = ds/dt a = dv/dt = d²s/dt²
Velocity is the first derivative of displacement with respect to time, and acceleration is the first derivative of velocity or the second derivative of displacement.
速度是位移对时间的一阶导数，加速度是速度对时间的一阶导数或位移对时间的二阶导数。
Conversely, to move from acceleration to velocity or from velocity to displacement, integrate with respect to time and add the constant of integration (initial condition). For example, v = ∫ a dt + u, and s = ∫ v dt + s₀.
反过来，从加速度求速度或从速度求位移，需要对时间积分并加上积分常数（初始条件）。例如，v = ∫ a dt + u，以及 s = ∫ v dt + s₀。
AQA questions might give an expression for a or v in terms of t and ask for s or the distance travelled. Distinguish between displacement (vector area) and distance (scalar total). To find distance when velocity changes sign, integrate the absolute value or split the motion at times when v = 0.
AQA的考题可能会给出用 t 表示的 a 或 v，要求求 s 或路程。要注意区分位移（代数量）和路程（标量总量）。若速度改变符号，求路程需要对绝对值积分或在 v = 0 的时刻拆分运动。
7. Projectile Motion: Horizontal and Vertical Components | 抛体运动：水平与竖直分量
A projectile is a particle moving freely under gravity, with the only acceleration being g acting vertically downwards. The key technique is to resolve the initial velocity into horizontal and vertical components and treat the two directions independently.
抛体是指仅在重力作用下自由运动的质点，其加速度仅为竖直向下的 g。关键技巧是将初速度分解为水平和竖直分量，并独立处理这两个方向。
If a particle is projected with speed U at an angle θ to the horizontal:
如果一个物体以速率 U、与水平面成 θ 角抛出：
Horizontal velocity: vₓ = U cosθ (constant)
Vertical velocity: vᵧ = U sinθ − g t
Horizontal displacement: x = (U cosθ) t
Vertical displacement: y = (U sinθ) t − ½ g t²
The horizontal motion is uniform because there is no horizontal acceleration. The vertical motion is uniformly accelerated with acceleration −g (assuming upward positive).
水平方向没有加速度，因此是匀速运动；竖直方向则是加速度为 −g 的匀加速运动（假设向上为正）。
Time of flight T = (2U sinθ)/g, maximum height H = (U² sin²θ)/(2g), and horizontal range R = (U² sin 2θ)/g. These formulas are derived by setting y = 0 or vᵧ = 0.
飞行时间 T = (2U sinθ)/g，最大高度 H = (U² sin²θ)/(2g)，水平射程 R = (U² sin 2θ)/g。这些公式可通过令 y = 0 或 vᵧ = 0 推导得出。
8. Vector Equations of Motion in Two Dimensions | 二维运动向量方程
In AQA exam questions, motion is often expressed using vectors. The position vector r, velocity v and acceleration a are written in terms of unit vectors i and j (horizontal and vertical directions).
在AQA考题中，运动经常用向量表示。位置矢量 r、速度 v 和加速度 a 用单位向量 i 和 j（水平和竖直方向）表示。
Differentiation and integration are performed component-wise. For example, if r = (3t²)i + (2t)j, then v = dr/dt = (6t)i + 2j and a = dv/dt = 6i. Conversely, given a constant acceleration a = 4i − gj, integrate with respect to t using initial velocity u.
微积分运算分别对各个分量进行。例如，若 r = (3t²)i + (2t)j，则 v = dr/dt = (6t)i + 2j，a = dv/dt = 6i。反之，若给定恒定加速度 a = 4i − gj，则结合初速度 u 对 t 积分即可。
When working with vector kinematics, you can find the speed (magnitude of velocity) at any time by taking the modulus, and the distance travelled is found by integrating the speed over time.
在处理向量运动学时，可通过求模得到任意时刻的速率，而路程则通过速率对时间积分求得。
9. Connected Particles and Pulley Scenarios | 连接体与滑轮情境
Although kinematics often focuses on single-particle motion, AQA may combine constant acceleration equations with connected particles. Here, you first use dynamics (F = ma) to find a common acceleration, then apply SUVAT to describe their motion.
虽然运动学通常关注单个质点，但AQA可能将匀加速方程与连接体结合。在这种情况下，你首先用动力学 (F = ma) 求出共同加速度，然后应用 SUVAT 描述它们的运动。
For example, two particles connected by a light inextensible string over a smooth pulley will have the same magnitude of acceleration. Once acceleration is found, the kinematic equations can predict speed after a given distance or time.
例如，两个物体通过轻质不可伸长的绳跨过光滑滑轮相连，它们的加速度大小相同。求出加速度后，运动学方程就能预测经过某段距离或时间后的速度。
Always ensure the kinematic portion uses the correct sign and magnitude of a, and remember that if the string breaks, each particle then moves under its own resulting force and possibly a new acceleration.
务必确保运动学部分使用了正确的加速度符号和大小，并记住如果绳子断开，之后每个物体将在各自的合外力下运动，可能产生新的加速度。
10. Common Pitfalls and Exam Tips | 常见错误与考试技巧
One common mistake is confusing displacement and distance. A particle may move back and forth; displacement may be zero while the distance travelled is not. Check whether the question asks for ‘distance’ or ‘displacement’.
一个常见错误是混淆位移和路程。物体可能来回运动，位移可能为零而路程不为零。要检查题目要求的是“路程”还是“位移”。
Another error is misusing SUVAT when acceleration is not constant. Always verify that a is constant over the whole motion before applying the equations; otherwise, switch to calculus.
另一个错误是在加速度不恒定时误用 SUVAT。在使用这些方程之前，务必确认在整个运动过程中加速度恒定；否则应改用微积分。
In projectile questions, forgetting to resolve the initial speed into components or using cos for the vertical component is a frequent slip. Carefully label the angle and draw a diagram. Also ensure g is used with the correct sign relative to your chosen positive direction.
在抛体问题中，忘记把初速度分解为分量，或者把竖直分量错用 cos 是常见的失误。务必仔细标示角度并画出示意图。同时要确保 g 的符号与所选的正方向一致。
For calculus problems, always find the constant of integration from given conditions. When calculating distance from a velocity-time graph or from a velocity function, divide the motion into intervals where v changes sign.
对于微积分问题，务必从所给条件中求出积分常数。根据速度-时间图或速度函数计算路程时，应将运动按速度改变符号的时刻分段。
Finally, practise past paper questions under timed conditions, and always present your reasoning clearly. A clear diagram and consistent vector notation can prevent many simple errors.
最后，要多在限定时间内练习往年真题，并始终清晰地呈现你的推理过程。清晰的示意图和一致的矢量符号可以避免许多简单错误。
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