Kinematics: Essential Revision Notes | 运动学：考点精讲

📚 Kinematics: Essential Revision Notes | 运动学：考点精讲

Kinematics is the study of motion without considering its causes. In the IGCSE WJEC Mathematics syllabus, you are expected to interpret and analyse motion using graphs, apply the constant acceleration equations (SUVAT), and solve problems involving displacement, velocity and acceleration. This article brings together all the key points you need, from definitions to worked examples, to help you revise efficiently.

运动学研究的是物体运动本身，而不涉及造成运动的原因。在 IGCSE WJEC 数学大纲中，你需要通过图像解释和分析运动，应用匀加速运动方程（SUVAT），并解决涉及位移、速度和加速度的问题。本文汇集了所有关键考点，从定义到典型例题，帮助你高效复习。

1. Scalar and Vector Quantities | 标量与矢量

In kinematics, we distinguish between scalars (magnitude only) and vectors (magnitude and direction). Distance and speed are scalars; displacement, velocity and acceleration are vectors. Always pay attention to direction when dealing with vector quantities, as it affects sign and interpretation.

在运动学中，我们区分标量（只有大小）和矢量（既有大小又有方向）。路程和速率是标量；位移、速度和加速度是矢量。处理矢量时一定要关注方向，因为它会影响正负号和物理意义。

2. Displacement, Velocity and Acceleration | 位移、速度与加速度

Displacement (s) is the change in position of an object from a reference point, measured in metres (m). Velocity (v) is the rate of change of displacement, given by v = Δs/Δt, measured in m/s. Acceleration (a) is the rate of change of velocity, given by a = Δv/Δt, measured in m/s². Deceleration is simply negative acceleration.

位移（s）是物体相对于参考点的位置变化，单位是米（m）。速度（v）是位移的变化率，即 v = Δs/Δt，单位是 m/s。加速度（a）是速度的变化率，即 a = Δv/Δt，单位是 m/s²。减速就是负的加速度。

When motion is in a straight line, these quantities can be assigned positive and negative signs according to a chosen direction. A negative velocity means the object is moving opposite to the positive direction, while a negative acceleration can indicate slowing down if velocity is positive, or speeding up if velocity is negative.

当运动沿直线发生时，可以根据选定的正方向为这些量赋予正负号。负速度表示物体运动方向与正方向相反；而负加速度在速度为正时表示减速，在速度为负时则表示加速。

3. Constant Acceleration Equations (SUVAT) | 匀加速运动方程 (SUVAT)

For motion in a straight line with constant acceleration, you can use the SUVAT equations. The five variables are s (displacement), u (initial velocity), v (final velocity), a (acceleration) and t (time). You need three known quantities to find a fourth. Learn these four equations thoroughly:

对于加速度恒定的直线运动，可以使用 SUVAT 方程。五个变量是 s（位移）、u（初速度）、v（末速度）、a（加速度）和 t（时间）。你需要已知三个量来求出第四个。请熟记以下四个方程：

v = u + at

s = ut + ½at²

v² = u² + 2as

s = ½(u + v)t

Always check that the acceleration is constant before applying these formulas. If the acceleration changes, break the motion into segments where it is constant, and apply SUVAT to each segment separately.

使用这些公式前，一定要确认加速度恒定。如果加速度发生变化，应将运动划分为加速度恒定的不同阶段，然后对每一段分别应用 SUVAT 方程。

4. Choosing the Right Equation | 正确选择方程

Each SUVAT equation is missing one variable: v = u + at misses s; s = ut + ½at² misses v; v² = u² + 2as misses t; s = ½(u+v)t misses a. Identify which quantity you need and which you don’t know, then select the equation that ignores the unknown variable.

每个 SUVAT 方程都缺少一个变量：v = u + at 不含 s；s = ut + ½at² 不含 v；v² = u² + 2as 不含 t；s = ½(u+v)t 不含 a。先确定需要求哪个量、已知哪些量，未知量是什么，然后选择不含那个未知量的方程。

5. Displacement–Time Graphs | 位移–时间图像

A displacement–time graph shows how displacement changes with time. The slope (gradient) of the graph represents velocity. A straight line means constant velocity; a curved line means changing velocity (acceleration). A horizontal line indicates the object is stationary.

位移–时间图像展示位移随时间的变化规律。图像的斜率表示速度。直线意味着速度恒定；曲线意味着速度变化（有加速度）。水平线表示物体静止。

The instantaneous velocity is found by drawing a tangent to the curve and calculating its gradient. Average velocity between two points is the chord gradient: (change in displacement)/(change in time).

瞬时速度可以通过在曲线上画切线并计算其斜率得到。两点之间的平均速度是弦的斜率：(位移变化量)/(时间变化量)。

6. Velocity–Time Graphs | 速度–时间图像

A velocity–time graph is one of the most important tools in kinematics. The gradient gives acceleration, and the area under the graph gives displacement. A horizontal line means constant velocity; a sloping straight line means constant acceleration.

速度–时间图像是运动学中最重要的工具之一。图像的斜率表示加速度，图像下的面积表示位移。水平线意味着速度恒定；倾斜直线意味着加速度恒定。

To find displacement from a v-t graph, calculate the area between the graph line and the time axis. Areas above the axis are positive displacement; areas below are negative. If the graph crosses the axis, total distance travelled is the sum of the absolute areas, while net displacement is the signed sum.

要从 v-t 图求位移，计算图线与时间轴之间的面积。轴上方面积为正位移，下方面积为负位移。如果图线穿过时间轴，则运动的总路程是各块面积的绝对值之和，而净位移是带符号的面积之和。

7. Acceleration–Time Graphs | 加速度–时间图像

An acceleration–time graph shows how acceleration varies with time. The area under an a-t graph gives the change in velocity (Δv). If the graph lies entirely above the axis, velocity increases; below the axis, velocity decreases.

加速度–时间图像展示加速度随时间的变化。a-t 图下的面积表示速度的变化量 (Δv)。如果图像完全在轴上方，速度增加；在轴下方则速度减小。

In many IGCSE problems, acceleration is constant, so the a-t graph appears as a horizontal line. Sudden changes in acceleration produce step-like graphs. Combine a-t information with initial velocity to construct the v-t graph step by step.

在许多 IGCSE 问题中，加速度是恒定的，因此 a-t 图表现为一条水平线。加速度的突变会产生阶梯状图像。结合 a-t 图的信息和初速度，可以逐步构造出 v-t 图。

8. Interpreting Motion from Graphs | 通过图像解读运动

A typical exam question gives a velocity–time graph and asks you to describe the motion in each section. You should mention whether the object is accelerating, decelerating, moving with constant velocity, or stationary. Also state the direction of motion (positive or negative) and calculate accelerations and distances travelled.

典型的考试题目会给出速度–时间图像，要求描述每一阶段的运动情况。你应该提到物体是在加速、减速、匀速还是静止，并说明运动方向（正或负），同时计算加速度和行进的路程。

For example, a v-t graph might show an initial sloping line upward (acceleration from rest), then a horizontal line (constant velocity), followed by a sloping line downward to zero (deceleration to rest), and finally a horizontal line at zero (at rest). Breaking the graph into time intervals makes analysis systematic.

例如，一幅 v-t 图可能先出现一条向上的斜线（从静止开始加速），然后是一条水平线（匀速），接着是一条向下斜线直至零（减速到静止），最后是位于零的水平线（静止）。将图像按时间段拆分可以使分析变得系统化。

9. Free Fall and Vertical Motion | 自由落体与竖直运动

Objects moving under gravity near the Earth’s surface experience a constant downward acceleration of g = 9.8 m/s². You can treat vertical motion with SUVAT, using a = ±9.8 m/s² depending on the sign convention. Upward is usually taken as positive, making a = –9.8 m/s² for an object thrown upwards.

在地球表面附近，受重力作用的物体会获得一个向下的恒定加速度，g = 9.8 m/s²。你可以使用 SUVAT 方程来处理竖直运动，根据正方向约定取 a = ±9.8 m/s²。通常取向上为正方向，这时上抛物体的加速度 a = –9.8 m/s²。

Key points for an object projected upward: at the highest point, v = 0 momentarily; the time to rise equals the time to fall back to the same level; the speed at any height is the same on the way up and down, but velocity direction is reversed.

竖直上抛物体的要点：在最高点瞬间 v = 0；上升到最高点的时间等于落回同一水平面的时间；同一高度处上升和下降的速率相同，但速度方向相反。

10. Solving Problems Involving Two Moving Objects | 涉及两个运动物体的题目

When two objects move along the same line, you may need to find when and where they meet, or when one overtakes the other. Write SUVAT expressions for each object’s displacement from a common origin. Set their displacements equal to find the meeting time, then substitute back to find position. Always define direction clearly.

当两个物体在同一直线上运动时，可能需要求它们何时何地相遇，或一个何时追及另一个。以共同原点为参考，分别写出每个物体的位移表达式。令两者位移相等求出相遇时间，再代回求得位置。务必明确设定正方向。

If the objects start at different times, make sure to adjust t appropriately, or use separate time variables. Draw a clear diagram showing initial positions, velocities and accelerations. This helps avoid sign errors.

如果两物体开始运动的时刻不同，要适当调整 t，或使用不同的时间变量。画一个清晰的示意图，标出初始位置、速度和加速度，这有助于避免符号错误。

11. Common Mistakes and How to Avoid Them | 常见错误及应对方法

Many students confuse displacement and distance, especially when velocity changes sign. Remember: distance is the total path length, while displacement is the straight-line change in position. In a v-t graph, area above the axis adds to positive displacement, area below subtracts. Total distance = sum of absolute areas.

许多学生混淆位移和路程，尤其是速度改变符号时。记住：路程是运动轨迹的总长度，而位移是位置变化的直线距离。在 v-t 图中，轴上方面积增加正位移，下方面积减去。总路程 = 各块面积绝对值之和。

Another frequent error is using the wrong SUVAT equation. Always list knowns and unknowns before choosing. Also, check units: convert km/h to m/s by dividing by 3.6, and minutes to seconds. Never mix units such as km and seconds without conversion.

另一个常见错误是选错 SUVAT 方程。选择之前一定要列出已知量和未知量。此外，要检查单位：将 km/h 转换为 m/s 需除以 3.6，分钟转换为秒。未经换算就混合使用 km 和秒等单位是绝对不可以的。

12. Worked Example (Typical Exam Question) | 典型例题精析

Question: A car accelerates uniformly from rest at 2 m/s² for 10 seconds, then maintains a constant velocity for 20 seconds, and finally decelerates uniformly at 4 m/s² until it stops. Find the total distance travelled.

题目：一辆汽车从静止开始以 2 m/s² 匀加速行驶 10 秒，然后匀速行驶 20 秒，最后以 4 m/s² 匀减速直至停下。求总路程。

Solution (Step 1 – Acceleration phase): u = 0, a = 2, t = 10. Use s = ut + ½at² → s₁ = 0 + ½ × 2 × 10² = 100 m. Final velocity v = u + at = 0 + 2×10 = 20 m/s.

解（第一步——加速阶段）： u = 0, a = 2, t = 10。用 s = ut + ½at² → s₁ = 0 + ½ × 2 × 10² = 100 m。末速度 v = u + at = 0 + 2×10 = 20 m/s。

Step 2 – Constant velocity phase: v = 20 m/s, t = 20 s. s₂ = v × t = 20 × 20 = 400 m.

第二步——匀速阶段： v = 20 m/s，t = 20 s。s₂ = v × t = 20 × 20 = 400 m。

Step 3 – Deceleration phase: u = 20, v = 0, a = –4. Use v² = u² + 2as → 0 = 20² + 2×(–4)×s₃ → 0 = 400 – 8s₃ → s₃ = 50 m. Total distance = 100 + 400 + 50 = 550 m.

第三步——减速阶段： u = 20，v = 0，a = –4。用 v² = u² + 2as → 0 = 20² + 2×(–4)×s₃ → 0 = 400 – 8s₃ → s₃ = 50 m。总路程 = 100 + 400 + 50 = 550 m。

This example shows how breaking the problem into constant-acceleration segments makes it manageable. Always draw a velocity–time sketch if it helps visualise the stages.

这个例题表明，将题目拆分为恒加速阶段可以使问题化繁为简。如果画一个速度–时间的草图有助于直观展示各个阶段，那就尽管画。

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