📚 A-Level Edexcel Maths: Kinematics Key Concepts | A-Level Edexcel 数学：运动学考点精讲
Kinematics is a cornerstone of the Mechanics component in Edexcel A-Level Mathematics. It focuses on describing how objects move, using vectors, SUVAT equations, calculus-based methods for varying acceleration, and graphical interpretations. A solid grasp of displacement, velocity, and acceleration — both as scalar magnitudes and directed vector quantities — is essential for tackling everything from simple straight-line motion to two-dimensional projectile problems. This revision guide brings together the must-know concepts, worked patterns, and exam tips to build your confidence.
运动学是Edexcel A-Level数学力学部分的核心支柱。它专注于描述物体如何运动，运用向量、SUVAT公式、基于微积分的变加速方法以及图像解析。牢固掌握位移、速度和加速度——无论是作为标量大小还是具有方向的向量——对于解决从简单直线运动到二维抛体问题都至关重要。本复习指南汇集了必知概念、解题模式和考试技巧，助你建立信心。
1. Core Quantities: Displacement, Velocity and Acceleration | 核心量：位移、速度和加速度
Displacement (s) is a vector that gives the change in position relative to an origin. 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. When motion is along a straight line, the sign convention (positive direction) must be defined clearly. In vector terms for two dimensions, we write r = xi + yj, v = dr/dt, a = dv/dt. For constant acceleration, the SUVAT equations connect these quantities without vectors in one-dimensional cases.
位移(s)是向量，表示相对于原点的位置变化。速度(v)是位移随时间的变化率，也是向量。加速度(a)是速度的变化率。当运动沿直线进行时，必须明确符号约定（正方向）。在二维向量形式中，可写为r = xi + yj，v = dr/dt，a = dv/dt。对于匀加速直线运动，SUVAT公式可以在不出现向量的情况下联系这些量。
2. The SUVAT Equations for Constant Acceleration | 匀加速运动的SUVAT公式
When acceleration is constant, five equations link the kinematic variables. They are derived assuming motion starts at time t=0, with initial velocity u, final velocity v, constant acceleration a and displacement s. The equations are:
当加速度恒定时，五个公式将运动学变量联系起来。它们的推导假设运动从时间 t=0 开始，初速度为 u，末速度为 v，恒定加速度为 a，位移为 s。公式如下：
| Equation | Notes |
|---|---|
| v = u + at | Missing s; useful for finding final velocity after time t |
| s = (u + v)/2 × t | Missing a; average velocity × time |
| s = ut + ½ at² | Missing v; commonly used when distance is required |
| s = vt − ½ at² | Missing u; less common but equally valid |
| v² = u² + 2as | Missing t; used when time is not involved |
Each SUVAT equation is used when three of the five quantities are known and a fourth is to be found. Always convert units to be consistent (e.g., metres and seconds) and choose a positive direction before substituting values.
每个SUVAT公式在已知五个量中的三个、欲求第四个时使用。务必使单位一致（如米和秒），并在代入数值前选定正方向。
3. Strategy: Selecting the Right SUVAT Equation | 策略：选择合适的SUVAT公式
To avoid errors, list the known quantities (s, u, v, a, t) and identify the one that is not required and the one you need to find. Then select the equation that excludes the irrelevant variable. For example, a question gives u=10 m/s, a=2 m/s², t=4 s and asks for s: you are missing v, so use s = ut + ½ at². If a ball is dropped from rest (u=0) from height h, you know s=−h (if upward is positive), a=−g, and you need v after falling, use v² = u² + 2as.
为了避免错误，列出已知量(s, u, v, a, t)，找出不需要的量和你需要求的量。然后选择不含无关变量的方程。例如，题目给出 u=10 m/s, a=2 m/s², t=4 s 要求 s：缺少 v，因此使用 s = ut + ½ at²。如果一个球从静止(u=0)从高度 h 落下，以向上为正，则 s=−h, a=−g，求末速度 v，使用 v² = u² + 2as。
4. Variable Acceleration: Calculus Connections | 变加速：微积分联系
When acceleration varies with time, you cannot use SUVAT. Instead, use differentiation and integration. If displacement is given as a function of time, x(t) or s(t), then v = ds/dt and a = dv/dt = d²s/dt². Conversely, given a(t), integrate to find v(t): v = ∫ a dt + C, and integrate again to find s(t): s = ∫ v dt + D, where C and D are constants determined by initial conditions. Common exam tasks include finding maximum velocity (set a=0), distance travelled in a time interval (integrate |v|), or the time when the particle changes direction (v=0).
当加速度随时间变化时，不能使用SUVAT，而要运用微分和积分。若已知位移为时间的函数 x(t) 或 s(t)，则 v = ds/dt，a = dv/dt = d²s/dt²。反过来，已知 a(t)，积分得 v(t)：v = ∫ a dt + C，再积分得 s(t)：s = ∫ v dt + D，其中 C 和 D 由初始条件确定。常见考题包括求最大速度（令 a=0），某时段内的路程（积分 |v|），或质点改变方向的时刻（v=0）。
5. Displacement–Time and Velocity–Time Graphs | 位移-时间与速度-时间图像
Graphs provide a visual way to interpret motion. On a displacement–time graph, the gradient at any point gives the velocity. A straight line implies constant velocity; a curve implies acceleration. On a velocity–time graph, the gradient gives acceleration, and the area under the graph between two times gives the displacement. For constant acceleration, the v–t graph is a straight line, and the area is a trapezium, leading directly to s = (u+v)/2 × t. Key skills include sketching, interpreting, and calculating areas formed by rectangles, triangles and trapeziums.
图像提供了理解运动的直观方式。在位移-时间图像上，任意点的斜率给出速度。直线表示匀速，曲线表示加速。在速度-时间图像上，斜率给出加速度，两时刻间图像下方的面积表示位移。对于匀加速运动，v-t 图是一条直线，面积是梯形，直接导出 s = (u+v)/2 × t。核心技能包括绘图、解读以及计算由矩形、三角形和梯形构成的面积。
6. Vertical Motion Under Gravity | 重力作用下的竖直运动
Objects moving freely near the Earth’s surface experience a constant downward acceleration g, usually taken as 9.8 m/s². When applying SUVAT, carefully assign signs. A typical convention is to take upward as positive, so a = −g. For a particle thrown upwards with speed u, the maximum height occurs when v=0: use v² = u² + 2as with s = H, giving H = u²/(2g). The time to reach the top is t = u/g. For a particle dropped from a height H, initial velocity u=0, and velocity just before impact is found from v² = 2gH. Symmetry means the time to go up equals the time to come down back to the same level.
在地表附近自由运动的物体受到恒定的向下加速度 g，通常取 9.8 m/s²。使用SUVAT时，需仔细分配符号。常见约定是向上为正，因此 a = −g。对于以初速 u 上抛的质点，最高点发生在 v=0 时：利用 v² = u² + 2as，令 s = H，得 H = u²/(2g)。到达最高点的时间 t = u/g。对从高度 H 自由落下的质点，初速 u=0，触地前的速度可由 v² = 2gH 求得。对称性意味着上升到最高点的时间等于从该高度落回出发点的时间。
7. Projectile Motion: Two-Dimensional Motion | 抛体运动：二维运动
Projectile motion can be split into independent horizontal and vertical components. The horizontal component has constant velocity (ux = u cos θ, ax = 0). The vertical component has constant downward acceleration g (uy = u sin θ, ay = −g when upward is positive). The trajectory is a parabola. Time of flight, range and maximum height are found using SUVAT on the vertical motion. For a projectile launched from the ground and landing at the same level, time of flight T = 2u sin θ / g, range R = u² sin 2θ / g, and maximum height H = u² sin² θ / (2g). Vector methods can unify these by writing r(t) = (u cos θ t) i + (u sin θ t − ½ g t²) j.
抛体运动可分解为独立的水平分量和竖直分量。水平分量为匀速运动(ux = u cos θ, ax = 0)。竖直分量具有向下的恒定加速度 g（当向上为正时，uy = u sin θ, ay = −g）。轨迹为抛物线。飞行时间、射程和最大高度通过对竖直运动使用SUVAT求得。对于从地面抛出并落回同一水平面的抛体，飞行时间 T = 2u sin θ / g，射程 R = u² sin 2θ / g，最大高度 H = u² sin² θ / (2g)。向量方法可以统一表示为 r(t) = (u cos θ t)i + (u sin θ t − ½ g t²)j。
8. Calculus with Vectors for Variable Acceleration | 变加速运动的向量微积分
In vector form, position r(t), velocity v(t) = dr/dt, and acceleration a(t) = dv/dt. If acceleration is given as a vector function of time, integrate componentwise to find velocity, and again to find displacement. For example, given a = 6t i + 4 j, integrate to get v = (3t² + C1)i + (4t + C2)j, using initial velocity to find constants. The speed is the magnitude |v|, and the direction is given by the unit vector v/|v|. To find when the particle is moving parallel to a vector, set the ratio of the velocity components equal to the direction ratio of that vector.
在向量形式下，位置 r(t)、速度 v(t) = dr/dt 和加速度 a(t) = dv/dt。若加速度以时间的向量函数给出，可对每个分量积分求得速度，再次积分求得位移。例如，给定 a = 6ti + 4j，积分得 v = (3t² + C₁)i + (4t + C₂)j，并利用初速度确定常数。速率为 |v|，方向由单位向量 v/|v| 给出。欲求质点何时与某向量平行，可令速度分量之比等于该向量的方向比。
9. Using Calculus to Determine Peak Values | 用微积分求解极值
In variable acceleration problems, the maximum or minimum displacement, velocity or speed often appears. To find maximum speed from v(t), differentiate speed squared or simply solve |v|’ = 0. However, for one-dimensional motion, setting velocity equal to zero often gives turning points for displacement. For a velocity function v = 12t − 3t², the particle changes direction when v=0, giving t=0 and t=4. The distance travelled requires integrating |v| over the interval. Maximum height for vertical projectile can also be found by setting vertical velocity component vy = 0.
在变加速问题中，常需求位移、速度或速率的极大或极小值。要从 v(t) 求最大速率，可对速度大小的平方求导或直接解 |v|’ = 0。但对一维运动，令速度为零通常给出位移的转折点。对于速度函数 v = 12t − 3t²，粒子在 v=0 时改变方向，得 t=0 和 t=4。路程需在区间上积分 |v|。竖直抛体的最大高度也可通过令竖直速度分量 vy = 0 求得。
10. Connected Particles and Kinematic Constraints | 连接体与运动学约束
When two particles are connected by a light inextensible string over a pulley or placed on a surface, their motions are linked. Kinematically, the speed of the connected particles is the same at any instant (if the string remains taut), and the magnitude of their accelerations is equal. For systems where one particle moves vertically and the other horizontally, you can use the same SUVAT concepts for each particle individually, but must relate their displacements and velocities through the string. In pulley problems, if one particle descends by distance h, the other rises by h, meaning the velocity magnitudes match.
当两个质点通过轻质不可伸长的绳子跨过滑轮或置于表面时，它们的运动相互关联。从运动学上看，连接质点的速率在任意时刻相同（若绳子保持绷紧），且它们的加速度大小相等。对于一质点竖直运动、另一质点水平运动的系统，可分别对每个质点运用SUVAT概念，但必须通过绳子关联它们的位移和速度。在滑轮问题中，若一个质点下降距离 h，另一个则上升 h，意味着速度大小匹配。
11. Common Mistakes and How to Avoid Them | 常见错误及其避免方法
Sign errors are the most frequent pitfall — always define a positive direction and stick to it. In SUVAT, check that the quantity you ignore is genuinely constant, especially in vertical motion where a = −g is constant only if gravity is the sole force. When integrating, never forget the constant of integration; use initial conditions to find it. In projectile problems, students often mix horizontal and vertical components or forget that horizontal velocity is constant. In graphical questions, confusing the gradient and the area under the
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