📚 IB WJEC Mathematics: Kinematics Key Points | IB WJEC 数学：运动学 考点精讲
Kinematics is the branch of mechanics that describes motion without considering its causes. In IB and WJEC mathematics, kinematics focuses on modelling objects as particles moving in one or two dimensions under constant acceleration. This article distils the essential definitions, formulas, graphical interpretations and problem‑solving techniques required for examination success.
运动学是力学中描述运动而无需考虑其成因的分支。在 IB 和 WJEC 数学中，运动学重点是将物体视作在常加速度下沿一维或二维运动的质点。本文提炼了应对考试所必须掌握的核心定义、公式、图像解读和解题技巧。
1. Displacement and Velocity | 位移与速度
Displacement is a vector quantity that measures the change in position of a particle from a fixed origin. It is usually denoted by s and measured in metres (m). Distance, on the other hand, is a scalar representing the total path length travelled. Always interpret displacement with its direction in one‑dimensional motion.
位移是描述质点相对于固定原点位置变化的矢量，通常用 s 表示，单位为米 (m)。而路程是标量，表示运动轨迹的总长度。在一维运动中，务必结合方向来理解位移的含义。
e velocity is total displacement divided by total time, whereas instantaneous velocity is the derivative of displacement. The magnitude of velocity is speed, a scalar.
速度是位移关于时间的变化率，即 v = ds/dt，单位为 m/s。平均速度是总位移除以总时间，瞬时速度则是位移的导数。速度的大小称为速率，是标量。
2. Acceleration | 加速度
a vector, measured in m/s². Constant acceleration is the most common model in these syllabuses, leading to the suvat equations.
加速度是速度关于时间的变化率，记作 a = dv/dt 或 a = d²s/dt²，也是矢量，单位为 m/s²。匀加速度是 IB 和 WJEC 课程中最常见的模型，由此引出 suvat 方程。
Deceleration (or retardation) simply means acceleration acting in the opposite direction to velocity. When solving problems, always assign a positive direction and treat acceleration as positive or negative accordingly.
减速（或负加速度）是指加速度方向与速度方向相反。解题时务必先规定正方向，加速度根据符号取正负。
3. The SUVAT Equations | 匀加速运动方程 (SUVAT)
For motion in a straight line with constant acceleration a, the following five equations link displacement s, initial velocity u, final velocity v, acceleration a, and time t. They are the cornerstone of kinematics problems.
在常加速度 a 的直线运动中，以下五个方程联系位移 s、初速度 u、末速度 v、加速度 a 和时间 t，是运动学问题的基石。
v = u + at
s = ut + ½ at²
v² = u² + 2as
s = ½ (u + v)t
s = vt − ½ at²
Memorising these identities is essential. The choice of equation depends on which three quantities are known and which one is required. Always write down the known variables before selecting the formula.
熟记这些等式至关重要。选择哪个方程取决于已知哪三个量、求哪一个量。套公式前务必先列出已知变量。
4. Free Fall and Acceleration Due to Gravity | 自由落体与重力加速度
When an object moves solely under gravity, its acceleration is g, taken as 9.8 m/s² or 9.81 m/s² downwards, unless the question specifies otherwise. In the absence of air resistance, all objects fall with the same acceleration.
物体仅在重力作用下运动时，其加速度为 g，通常取 9.8 m/s² 或 9.81 m/s²，方向向下（除非题目另有说明）。在忽略空气阻力的情况下，所有物体下落加速度相同。
n sign convention. At the highest point, the instantaneous velocity is zero, but acceleration remains g.
竖直抛体问题可直接使用 suvat 方程，根据规定的正方向令 a = −g 或 a = +g。在最高点时瞬时速度为零，但加速度仍为 g。
5. Displacement-Time Graphs | 位移-时间图
A displacement–time (s–t) graph plots displacement against time. The gradient of the tangent at any point gives the instantaneous velocity. A straight line indicates constant velocity; a curved line indicates changing velocity.
位移-时间 (s–t) 图描点得到位移随时间的变化。曲线上任一点切线的斜率即瞬时速度。直线表示匀速运动，曲线表示变速运动。
If the graph is a parabola, the motion involves constant acceleration. Be careful: the area under an s–t graph has no physical meaning for motion. Questions often ask for average velocity, which is the slope of the chord connecting two points.
若图像为抛物线，则运动含匀加速度。注意：位移-时间图下的面积在运动学中没有物理意义。题目常要求求平均速度，即连接两点的弦的斜率。
6. Velocity-Time Graphs | 速度-时间图
A velocity–time (v–t) graph is one of the most powerful tools in kinematics. The gradient gives acceleration; the area between the graph and the time axis gives displacement (with areas below the axis counting as negative).
速度-时间 (v–t) 图是运动学中最强大的工具之一。斜率表示加速度；图线与时间轴围成的面积表示位移（时间轴以下面积为负）。
For constant acceleration, the v– )t. Exam questions frequently require calculating displacement from a graph, so practise area splitting for shapes like triangles, rectangles and trapeziums.
匀加速运动的 v–t 图是一条直线，图下面积构成梯形，直接导出公式 s = ½ (u+v)t。考试中常要求根据图像计算位移，因此要多练习将图形分割为三角形、矩形和梯形来求面积。
7. Acceleration-Time Graphs | 加速度-时间图
An acceleration–time (a–t) graph plots acceleration against time. The area under the graph gives the change in velocity (Δv). For constant acceleration, the graph is a horizontal line.
加速度-时间 (a–t) 图描点得到加速度随时间的变化。图下面积表示速度变化量（Δv）。匀加速运动的 a–t 图像是一条水平线。
These graphs are less common in straight‑line kinematics but become useful when analysing motion with piecewise constant acceleration. Always note that the gradient of an a–t graph is jerk, a quantity rarely examined at this level.
这类图在直线运动学中不太常见，但分析分段匀加速运动时很有用。注意 a–t 图的斜率表示急动度，该量在此阶段极少考到。
8. Projectile Motion | 抛体运动
A projectile moves under constant vertical acceleration due to gravity and constant horizontal velocity (ignoring air resistance). The horizontal and vertical components of motion are independent. The resultant path is a parabola.
抛体在竖直方向受重力加速度影响作匀变速运动，水平方向速度不变（忽略空气阻力）。水平和竖直分量相互独立，合运动轨迹为抛物线。
and vertical (aᵧ = −g, vᵧ = u sin θ) directions. The vertical velocity at maximum height is zero, giving the time to peak as t = (u sin θ)/g.
典型问题包括求飞行时间、最大高度或水平射程。水平方向使用匀速度公式 (aₓ = 0, vₓ = u cos θ)，竖直方向使用匀加速公式 (aᵧ = −g, vᵧ = u sin θ)。最高点竖直速度为零，到达最高点的时间为 t = (u sin θ)/g。
Maximum height H = (u² sin² θ) / (2g)
Range R = (u² sin 2θ) / g
These derived formulas can be quoted unless the question demands a full derivation.
上述推导公式可直接引用，除非题目要求完整推导过程。
9. Relative Motion | 相对运动
Relative velocity describes the motion of one object as observed from another. If two particles A and B have velocities vₐ and vᵦ, the velocity of A relative to B is vₐ − vᵦ. This concept is important for collision‑avoidance problems and river‑crossing scenarios.
相对速度描述从一个物体观察到的另一个物体的运动。若质点 A、B 的速度分别为 vₐ 和 vᵦ，则 A 相对于 B 的速度为 vₐ − vᵦ。这一概念对避免碰撞问题和渡河情景十分关键。
Draw velocity vector diagrams to find resultant relative speed and direction. In one dimension, simply subtract the velocities, paying careful attention to signs representing direction.
作速度矢量图可求出相对速度的大小和方向。在一维情形中，直接相减并注意正负号代表的方向即可。
10. Calculus in Kinematics | 运动学中的微积分
s(t), differentiate to get velocity v(t) = ds/dt, and differentiate again for acceleration a(t) = dv/dt = d²s/dt².
当加速度非恒定时，位移、速度和加速度通过求导和积分联系。给定位移 s(t)，求导得速度 v(t) = ds/dt，再次求导得加速度 a(t) = dv/dt = d²s/dt²。
cement: s = ∫ v dt + D. Initial conditions are used to determine constants of integration.
反之，给定加速度 a(t)，积分得速度：v = ∫ a dt + C；再积分速度得位移：s = ∫ v dt + D。利用初始条件确定积分常数。
Problems often involve finding maximum displacement or velocity by setting the derivative equal to zero. Be comfortable with both indefinite and definite integrals for calculating changes over a time interval.
常见问题包括令导数为零求最大位移或速度。对于用定积分求某时间区间内的变化量，也应能熟练运用。
11. Problem‑Solving Strategy for 1D Kinematics | 一维运动学解题策略
A systematic approach drastically reduces errors:
- Draw a clear diagram and label the origin and positive direction.
- List the known quantities: s, u, v, a, t.
- Identify the required unknown.
- Choose the suvat equation that includes the knowns and the unknown.
- Substitute values and solve, checking that units are consistent.
- Interpret the result physically (e.g. negative displacement means opposite direction).
系统方法能大大降低出错率：
- 画清晰示意图，标出原点和正方向；
- 列出已知量：s、u、v、a、t；
- 找出待求未知量；
- 选择包含已知量与未知量的 suvat 方程；
- 代入数值求解，确保单位一致；
- 从物理意义解释结果（如负位移表示与正方向相反）。
12. Common Mistakes and Exam Tips | 常见错误与应试技巧
is, its velocity and displacement are negative. In vertical motion, using g = 9.8 often causes calculation slips; use exact values until the final answer.
小心符号错误：始终声明正方向并保持一致。物体运动与所设正方向相反时，其速度和位移均为负值。竖直运动中使用 g = 9.8 时常导致计算疏忽，建议保留精确值，最终一步再代入。
Many candidates confuse distance and displacement. If a particle changes direction in one‑dimensional motion, distance must be calculated by summing the absolute lengths of each segment, whereas displacement is the net change in position. Apply the correct area interpretation on velocity‑time graphs.
许多考生混淆路程和位移。如果质点在直线上改变方向，路程需将各段轨迹的绝对值相加，位移则是位置的净变化。在速度-时间图上要注意区分面积的正负。
For projectile motion, separate horizontal and vertical components at the start. Never mix them in a single suvat equation. Finally, check whether the problem involves particles meeting or overtaking, which requires setting their displacements equal at the same time.
对于抛体运动，一开始就要分解水平和竖直分量，切勿在同一个 suvat 方程中混用。最后，确认问题是否涉及质点相遇或追及，这需要令它们的位移在同一时刻相等。
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