📚 IGCSE CIE Mathematics: Kinematics Key Points | IGCSE CIE 数学：运动学考点精讲
Kinematics is the branch of mathematics that describes the motion of objects using displacement, velocity, and acceleration, without referring to the forces causing the motion. For IGCSE CIE Mathematics, you must be confident with constant acceleration equations (SUVAT), distance–time and velocity–time graphs, and vertical motion under gravity. This guide walks you through every essential concept, common pitfalls, and exam strategies to help you secure top marks.
运动学是数学中描述物体运动的分支，使用位移、速度和加速度，而不涉及引起运动的力。在 IGCSE CIE 数学中，你必须熟练掌握匀加速运动方程（SUVAT）、距离–时间图与速度–时间图，以及重力作用下的竖直运动。本文带你梳理所有重要概念、常见错误和应试策略，助你稳拿高分。
1. Basic Concepts of Kinematics | 运动学基本概念
Displacement is a vector quantity that measures the change in position of an object in a particular direction. Distance is a scalar that represents the total length of the path travelled without regard to direction. Similarly, velocity is a vector describing the rate of change of displacement, while speed is the scalar magnitude of velocity. Acceleration is the rate of change of velocity, and can be positive (speeding up in the positive direction) or negative (slowing down, often called deceleration).
位移是矢量，测量物体在特定方向上的位置变化。距离是标量，表示走过的路径总长度，不考虑方向。同样，速度是矢量，描述位移的变化率；速率是速度的大小，为标量。加速度是速度的变化率，可为正（正方向加速）或负（减速，通常称为减速度）。
In IGCSE problems, we usually consider motion in a straight line. We define a positive direction (e.g. rightwards or upwards) and assign signs to all vector quantities. A negative velocity means the object is moving opposite to the chosen positive direction; a negative acceleration means the object is decelerating or accelerating in the negative direction.
在 IGCSE 问题中，我们通常研究直线运动。我们定义一个正方向（如向右或向上），并给所有矢量赋上正负号。负速度表示物体沿所选正方向的反方向运动；负加速度表示物体正在减速，或沿负方向加速。
The standard units used are metres (m) for displacement, metres per second (m/s) for velocity, and metres per second squared (m/s²) for acceleration. Time is measured in seconds (s). Always check that units are consistent before substituting into formulas.
标准单位：位移用米（m），速度用米每秒（m/s），加速度用米每二次方秒（m/s²）。时间用秒（s）。在代入公式之前，务必确保所有单位一致。
2. The SUVAT Equations | 匀加速运动方程（SUVAT）
When an object moves with uniform (constant) acceleration in a straight line, five quantities are related: s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time taken. The four equations of motion, often called the SUVAT equations, allow us to find any two unknowns provided three of the quantities are known.
当物体沿直线做匀加速（加速度恒定）运动时，五个量相互关联：s = 位移，u = 初速度，v = 末速度，a = 加速度，t = 所用时间。这四个运动方程（常称 SUVAT 方程）使我们能在已知三个量的情况下，求出另外两个未知量。
| Symbol | Meaning (EN) | 含义 (CN) | SI Unit |
|---|---|---|---|
| s | displacement | 位移 | m |
| u | initial velocity | 初速度 | m/s |
| v | final velocity | 末速度 | m/s |
| a | acceleration | 加速度 | m/s² |
| t | time | 时间 | s |
① v = u + at
② s = ut + ½at²
③ v² = u² + 2as
④ s = ½(u + v)t
These equations are only valid when acceleration is constant. You must select the equation that contains the three known quantities and the one unknown you wish to find. For example, if you are given u, a, t and need v, use equation ①. If u, v, t are given and you need s, equation ④ is the quickest.
这些方程仅在加速度恒定时成立。你必须选择包含三个已知量和所求未知量的方程。比如，已知 u、a、t，求 v，则用方程 ①；已知 u、v、t，求 s，方程 ④ 最快。
A common trick is to list the five symbols, note the numerical values with their signs, and mark a question mark for the unknown. This helps you avoid choosing an equation that lacks a needed quantity.
常见的技巧是列出五个符号，标注数值和正负号，未知量打问号。这能帮你避免选错方程，避免方程中缺少所需量。
3. Distance–Time Graphs | 距离–时间图
A distance–time graph plots total distance travelled against time. Since distance is a scalar, the graph always slopes upwards or is horizontal; it never decreases. The gradient (slope) of a distance–time graph represents the speed of the object. A steeper slope means higher speed. A horizontal line means the object is stationary (speed = 0).
距离–时间图描绘物体走过的总距离随时间的变化。因为距离是标量，图线总向上倾斜或为水平线，从不下降。距离–时间图的斜率（坡度）表示物体的速率。坡度越陡，速率越高。水平线表示物体静止（速率为零）。
To find the speed at any point, calculate the gradient of the tangent if the graph is curved (changing speed), or the gradient of the straight‑line segment if the speed is constant. Since distance is always positive, the speed calculated is always non‑negative.
欲求某点的速率，若图线是曲线（速率变化），计算该点切线的斜率；若图线是直线段（恒定速率），直接计算该线段的斜率。因为距离始终为正，算出的速率非负。
Distance–time graphs cannot show the direction of motion. They only show how far the object has travelled in total. Therefore, they are less informative than displacement‑time graphs for vector motion, but they still appear in IGCSE questions.
距离–时间图无法展示运动方向，只显示物体总共走了多远。因此，对于矢量运动，它们不如位移–时间图信息丰富，但仍在 IGCSE 考题中出现。
4. Velocity–Time Graphs | 速度–时间图
A velocity–time graph shows how velocity (a vector) changes with time. The velocity axis has positive and negative values, allowing the graph to show direction changes. The gradient of a velocity–time graph gives the acceleration: a positive slope means acceleration in the positive direction, a negative slope means deceleration or acceleration in the negative direction. A horizontal line indicates constant velocity (zero acceleration).
速度–时间图展示速度（矢量）随时间的变化。速度轴包含正负值，可显示方向变化。速度–时间图的斜率给出加速度：正斜率表示正方向加速，负斜率表示减速或负方向加速。水平线代表速度恒定（加速度为零）。
The area under a velocity–time graph between two times represents the displacement (not distance) during that interval. If the graph dips below the time axis (negative velocity), the area counts as negative displacement. To find the total distance travelled, you must take the absolute value of each area section separately.
速度–时间图在某段时间内与时间轴围成的面积代表该时间段内的位移（不是距离）。若图线在时间轴下方（负速度），该部分面积计为负位移。要求总路程，必须分别取每一段面积的绝对值再相加。
These two properties—gradient gives acceleration, area gives displacement—are extremely powerful. Always annotate your graph with slopes and area breakdowns when solving problems. Many IGCSE questions ask you to find acceleration from a slope or distance/displacement from an area.
这两个性质——斜率求加速度，面积求位移——非常有用。解题时务必在图上标出斜率和面积划分。很多 IGCSE 题目要求你从斜率求加速度，或从面积求距离/位移。
5. Calculating Slope and Area from Graphs | 从图中计算斜率和面积
To compute the gradient of a straight line on any graph, choose two well‑separated points (x₁, y₁) and (x₂, y₂). The gradient is (y₂ − y₁) / (x₂ − x₁). For a velocity–time graph, this gradient is the acceleration. Ensure you read the axes units carefully; a common mistake is to mix seconds and minutes.
计算任意图中直线斜率时，选取两个相距较远的点 (x₁, y₁) 和 (x₂, y₂)，斜率为 (y₂ − y₁) / (x₂ − x₁)。对速度–时间图，此斜率就是加速度。务必仔细读取坐标轴单位，常见错误是混淆秒与分钟。
For a curved distance–time or velocity–time graph, you may need to draw a tangent at the point of interest and then find its gradient. Practice drawing accurate tangents with a ruler—marks are awarded for correct construction lines.
若距离–时间或速度–时间图为曲线，你可能需要在目标点画切线，再求切线斜率。用直尺练习画准确切线——作图辅助线正确可得步骤分。
To find area under a velocity–time graph, break the shape into rectangles, triangles, or trapeziums. Then sum the signed areas. For example, a triangle above the axis gives positive displacement; a triangle below gives negative displacement. Total distance = sum of absolute values.
求速度–时间图下的面积时，将形状分割为矩形、三角形或梯形，然后求代数和。例如，轴上方的三角形给出正位移；轴下方的三角形给出负位移。总路程 = 各部分面积的绝对值之和。
6. Motion Under Gravity | 重力作用下的运动
When an object moves vertically near the Earth’s surface and air resistance is neglected, it experiences a constant acceleration due to gravity, g. In IGCSE CIE Mathematics, you may use g = 10 m/s² (if stated in the question) or g = 9.8 m/s². The direction of g is vertically downwards.
物体在地表附近竖直运动且忽略空气阻力时，受到恒定的重力加速度 g。在 IGCSE CIE 数学中，若题目注明，可使用 g = 10 m/s² 或 g = 9.8 m/s²。g 的方向竖直向下。
Choose a sign convention carefully: usually, upwards is taken as positive. Then the acceleration a = −g (since it acts downwards). For an object thrown upwards, the velocity becomes zero at the highest point, but acceleration is still −g. The same SUVAT equations apply with a = −g.
谨慎选择正负号惯例：通常取向上为正，则加速度 a = −g（因其方向向下）。对于上抛物体，在最高点速度为零，但加速度仍为 −g。相同的 SUVAT 方程适用，其中 a = −g。
If an object is dropped from rest, u = 0, a = g (taking downwards as positive) or u = 0, a = −g (upwards positive). Common question types: finding maximum height, time of flight, speed on hitting the ground. Always identify which SUVAT equation avoids the need for the unknown time or displacement when appropriate.
若物体从静止下落，取向下为正则 u = 0, a = g；取向上为正则 u = 0, a = −g。常见题型：求最大高度、飞行时间、落地速度。解题时，识别哪个 SUVAT 方程无需未知时间或位移最为恰当。
7. Applying SUVAT: Step‑by‑Step Strategy | 应用 SUVAT：分步解题策略
Step 1: Read the problem and draw a simple diagram showing the positive direction, starting point, and any known vectors.
第1步： 阅读题目，画简图标明正方向、起点和已知矢量。
Step 2: Write down the five symbols s, u, v, a, t. Fill in the known values with correct signs. Put a question mark for the unknown(s) you need to find.
第2步： 写下五个符号 s, u, v, a, t。填入已知值并带有正负号。未知量打问号。
Step 3: Check that all units are consistent (metres, seconds). Convert if necessary (e.g. km/h to m/s).
第3步： 检查所有单位是否一致（米、秒）。必要时进行换算（例如 km/h 转 m/s）。
Step 4: Select the SUVAT equation that contains the three knowns and the one target unknown. Write it down.
第4步： 选择包含三个已知量和目标未知量的 SUVAT 方程，写出方程。
Step 5: Substitute the values and solve algebraically. Keep signs intact.
第5步： 代入数值并代数求解，保留正负号。
Step 6: Interpret the result physically. If you get a negative time, check your signs or discard the non‑physical root.
第6步： 物理意义解读。若得到负时间，检查正负号或舍去无物理意义的根。
This systematic method reduces errors and ensures you don’t accidentally use an equation that requires a quantity you don’t have. Practice it until it becomes automatic.
这套系统方法可减少错误，确保你不会误用需要未知量的方程。不断练习直至熟练自如。
8. Worked Example: Horizontal Motion | 实例精讲：水平运动
Problem: A car accelerates uniformly from rest at 3 m/s² for 8 seconds. Find (a) its final velocity, and (b) the distance travelled during this time.
题目： 一辆汽车从静止开始以 3 m/s² 匀加速行驶 8 秒。求 (a) 末速度，(b) 这段时间内行驶的距离。
Solution (a): List s = ?, u = 0, v = ?, a = 3, t = 8. To find v, use v = u + at. v = 0 + 3 × 8 = 24 m/s. The final velocity is 24 m/s.
解 (a)： 列出 s = ?, u = 0, v = ?, a = 3, t = 8。使用 v = u + at，v = 0 + 3 × 8 = 24 m/s。末速度为 24 m/s。
Solution (b): With u, a, t known, we can use s = ut + ½at². s = 0 × 8 + ½ × 3 × 8² = 0 + ½ × 3 × 64 = 96 m. Alternatively, using s = ½(u + v)t = ½(0 + 24)×8 = 96 m. The car travels 96 m.
解 (b)： 已知 u, a, t，可用 s = ut + ½at²。s = 0 × 8 + ½ × 3 × 8² = 96 m。亦可使用 s = ½(u + v)t = ½(0 + 24)×8 = 96 m。汽车行驶了 96 m。
9. Common Mistakes to Avoid | 常见错误避坑
Mistake 1: Forgetting to assign a positive direction and applying signs inconsistently, especially in vertical motion. Always state your sign convention at the start.
错误 1： 忘记设定正方向，符号运用不一致，尤其在竖直运动中。务必在开始时声明正负号约定。
Mistake 2: Mixing units, for instance using km/h with seconds. Convert everything to m and s before applying SUVAT.
错误 2： 单位混用，例如将 km/h 与秒一起使用。在应用 SUVAT 前，将所有量换算为米和秒。
Mistake 3: Using the wrong graph property: thinking the gradient of a distance–time graph gives acceleration, or that the area under a distance–time graph gives displacement. Remember: gradient of distance–time = speed; gradient of velocity–time = acceleration; area under velocity–time = displacement.
错误 3： 混淆图形性质：误以为距离–时间图的斜率代表加速度，或距离–时间图下的面积代表位移。
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