A-Level WJEC Maths: Kinematics Key Points Explained | A-Level WJEC 数学：运动学考点精讲

📚 A-Level WJEC Maths: Kinematics Key Points Explained | A-Level WJEC 数学：运动学考点精讲

Kinematics is the branch of mechanics that describes the motion of objects along a straight line, without reference to the forces that cause the motion. In WJEC A-Level Mathematics, kinematics appears in the applied unit and draws heavily on algebraic manipulation, graph interpretation, and calculus. Students must be confident moving between displacement, velocity, and acceleration, using SUVAT equations for constant acceleration, analysing motion graphs, and applying differentiation and integration to variable acceleration problems. This article sets out the essential knowledge and exam techniques required to master kinematics.

运动学是力学的一个分支，描述物体沿直线的运动，而不涉及引起运动的力。在 WJEC A-Level 数学中，运动学出现在应用单元，大量依赖代数运算、图像解读和微积分。学生必须能够自如地在位移、速度和加速度之间转换，对匀加速运动使用 SUVAT 方程，分析运动图像，并将微分和积分应用于变加速问题。本文梳理了掌握运动学所需的核心知识和考试技巧。

1. Displacement, Velocity and Acceleration | 位移、速度和加速度

Displacement (s) is a vector quantity measuring the distance of a particle from a fixed origin in a specific direction. Velocity (v) is the rate of change of displacement with respect to time, and acceleration (a) is the rate of change of velocity. The fundamental relationships are:

位移 (s) 是一个矢量，度量质点沿特定方向到固定原点的距离。速度 (v) 是位移对于时间的变化率，而加速度 (a) 是速度的变化率。基本关系式为：

v = ds/dt, a = dv/dt = d²s/dt²

Displacement is measured in metres (m), velocity in metres per second (m s⁻¹), and acceleration in metres per second squared (m s⁻²). In one-dimensional motion, a positive sign usually indicates movement to the right or upwards, while a negative sign indicates the opposite direction.

位移的单位是米 (m)，速度的单位是米每秒 (m s⁻¹)，加速度的单位是米每二次方秒 (m s⁻²)。在一维运动中，正号通常表示向右或向上运动，而负号表示相反的方向。

2. SUVAT Equations for Constant Acceleration | 匀加速运动公式

When acceleration is constant, five key equations link the quantities s (displacement), u (initial velocity), v (final velocity), a (acceleration), and t (time). These are known as the SUVAT equations:

当加速度恒定时，五个关键方程将位移 s、初速度 u、末速度 v、加速度 a 和时间 t 联系起来。这些方程称为 SUVAT 方程：

v = u + at

s = ut + ½at²

s = vt − ½at²

v² = u² + 2as

s = ½(u + v)t

Each equation omits one of the five quantities: the first omits s, the second omits v, the third omits u, the fourth omits t, and the fifth omits a. Choosing the correct equation depends on which variables are known and which is required.

每个方程都省略了五个量中的一个：第一个省略 s，第二个省略 v，第三个省略 u，第四个省略 t，第五个省略 a。选择正确的方程取决于已知哪些变量以及需要求出哪个变量。

3. Applying SUVAT – Problem Solving Strategy | 应用 SUVAT 解题策略

In an exam, you should always list the known values and the unknown before selecting an equation. A systematic approach reduces errors:

在考试中，你应在选择方程之前列出已知值和未知量。系统的方法可以减少错误：

Step 1: Write down s, u, v, a, t and assign known numerical values with signs.
步骤1：写下 s, u, v, a, t 并赋予带有符号的已知数值。
Step 2: Identify which quantity is not mentioned in the problem and which equation omits it.
步骤2：确定问题中没有提及哪个量，以及哪个方程省略了它。
Step 3: Substitute the values and solve the resulting equation. Check that the answer has the correct unit and sign.
步骤3：代入数值并解出方程。检查答案是否有正确的单位和符号。

For example, a car accelerates from rest at 3 m s⁻² for 8 seconds. To find the distance covered: known are u = 0, a = 3, t = 8; unknown is s; v is not needed so use s = ut + ½at² = 0 × 8 + ½ × 3 × 8² = 96 m.

例如，一辆汽车从静止以 3 m s⁻² 加速 8 秒。求行驶距离：已知 u = 0, a = 3, t = 8；未知 s；不需要 v，所以用 s = ut + ½at² = 0 × 8 + ½ × 3 × 8² = 96 m。

4. Vertical Motion under Gravity | 重力作用下的竖直运动

When a particle moves vertically near the Earth’s surface, the acceleration due to gravity is g = 9.8 m s⁻² downwards. It is essential to choose a consistent positive direction. If upward is positive, then a = −9.8 m s⁻². The same SUVAT equations apply.

当质点在地球表面附近竖直运动时，重力加速度为 g = 9.8 m s⁻²，方向向下。选择一致的正方向至关重要。如果向上为正，那么加速度 a = −9.8 m s⁻²。同样适用 SUVAT 方程。

For an object thrown upwards with initial speed 14.7 m s⁻¹, the time to reach maximum height is found using v = u + at with v = 0, u = 14.7, a = −9.8. This gives t = 1.5 s. The maximum height can then be found from s = ut + ½at².

对于以初速度 14.7 m s⁻¹ 向上抛出的物体，达到最大高度的时间可用 v = u + at 求解，其中 v = 0, u = 14.7, a = −9.8。得 t = 1.5 s。最大高度可由 s = ut + ½at² 求出。

5. Velocity-Time Graphs and Their Interpretation | 速度-时间图及其解读

A velocity-time (v–t) graph plots velocity on the vertical axis against time on the horizontal axis. The gradient of the graph at any point gives the acceleration. A straight sloping line represents constant acceleration, and a horizontal line represents constant velocity. The area between the graph and the time axis represents the displacement. Areas above the axis are positive displacement; areas below are negative.

速度-时间 (v–t) 图将速度置于纵轴，时间置于横轴。图线上任意一点的斜率给出加速度。倾斜直线表示匀加速，水平线表示匀速。图线与时间轴之间的面积表示位移。轴上方的面积为正位移，下方为负。

To calculate total distance travelled when the graph crosses the time axis, you must sum the absolute values of the areas. Do not simply integrate; split the graph into sections where velocity has a constant sign.

当图线穿越时间轴时，要计算总路程，必须将各部分面积的绝对值相加。不能简单积分；应将图线按速度符号不变分段处理。

6. Displacement-Time and Acceleration-Time Graphs | 位移-时间图和加速度-时间图

A displacement-time (s–t) graph shows how displacement varies with time. Its gradient gives the velocity. A curve with increasing gradient indicates acceleration, while a straight line corresponds to constant velocity. A horizontal s–t graph means the particle is at rest.

位移-时间 (s–t) 图展示了位移随时间的变化。其斜率给出速度。斜率递增的曲线表示加速，直线对应于匀速。水平的 s–t 图表示质点静止。

An acceleration-time (a–t) graph is less common at this level but illustrates when acceleration is constant or varying. The area under an a–t graph over a time interval represents the change in velocity during that interval. When acceleration is constant, the graph is a horizontal line, and the area is a simple rectangle.

加速度-时间 (a–t) 图在这个阶段较少出现，但可展示加速度是恒定还是变化的。在某个时间段内 a–t 图下的面积表示该段时间内速度的变化量。当加速度恒定时，图线是一条水平线，面积为简单的矩形。

7. Using Differentiation to Find Velocity and Acceleration | 用微分求速度和加速度

When displacement is given as a function of time, s(t), velocity is the first derivative and acceleration is the second derivative. This is particularly relevant when acceleration is not constant. For example, if s = t³ − 6t² + 9t, then:

当位移表示为时间的函数 s(t) 时，速度是一阶导数，加速度是二阶导数。这在加速度不恒定时尤其重要。例如，若 s = t³ − 6t² + 9t，则：

v = ds/dt = 3t² − 12t + 9

a = dv/dt = 6t − 12

You may be asked to find the times when the particle is at rest (v = 0) or when the acceleration is zero. Always differentiate carefully and check your algebra.

你可能会被要求找出质点静止的时刻 (v = 0) 或加速度为零的时刻。仔细微分并检查你的代数。

8. Using Integration to Recover Displacement and Velocity | 用积分求回位移和速度

Conversely, if acceleration is given as a function of time, a(t), integration yields velocity, and a second integration gives displacement. Initial conditions are needed to find the constants of integration. For instance, given a = 4t − 2, with v = 5 when t = 0 and s = 3 when t = 0:

相反地，如果加速度以时间的函数 a(t) 给出，积分可得速度，二次积分可得位移。需要初始条件来确定积分常数。例如，已知 a = 4t − 2，且在 t = 0 时 v = 5，t = 0 时 s = 3：

v = ∫(4t − 2) dt = 2t² − 2t + C

Using t = 0, v = 5 gives C = 5, so v = 2t² − 2t + 5. Then

代入 t = 0, v = 5 得 C = 5，所以 v = 2t² − 2t + 5。接着

s = ∫(2t² − 2t + 5) dt = (2/3)t³ − t² + 5t + D

Using s = 3 at t = 0 gives D = 3. Displacement as a function of time is now fully determined.

利用 t = 0, s = 3 得 D = 3。位移作为时间的函数现已完全确定。

9. Distance Travelled versus Displacement | 路程与位移的区别

Displacement is the net change in position, while distance is the total length of the path travelled. When velocity changes sign, the particle reverses direction, and the total distance must be calculated by integrating the absolute value of velocity or by working out the separate displacements in each time interval and adding their magnitudes. Many WJEC exam questions specifically ask for the distance travelled between two times, and marks are lost if you merely compute the final displacement.

位移是位置的净变化，而路程是运动路径的总长度。当速度改变符号时，质点反转方向，总路程须通过积分速度的绝对值或将每个时间段内的位移大小相加来计算。许多 WJEC 考题明确要求计算一段时间内通过的路程，如果只计算末位移就会失分。

To find the distance when velocity is given by v = t² − 4t + 3, first solve v = 0 to identify turning points, then evaluate ∫|v| dt over the interval. Never ignore sign changes.

当速度由 v = t² − 4t + 3 给出时，先解 v = 0 找出转向点，再在区间上求 ∫|v| dt。切勿忽视符号变化。

10. Common Pitfalls and Exam Tips | 常见错误与考试技巧

Direction signs: Always define a positive direction at the start and keep it consistent throughout the question. Misplaced signs lead to completely wrong answers.
方向符号：在解题开始时始终定义正方向，并在全题中保持一致。符号错误会导致答案全错。
Units: Convert all quantities to SI units before substituting into formulas. Mixed units (e.g. km and m) are a frequent source of error.
单位：在代入公式前将所有量转换为国际单位。混合单位（如千米和米）是常见的错误来源。
SUVAT selection: Do not guess the equation. Cross out the variable that is neither given nor required, then use the equation that omits it.
SUVAT 选择：不要猜测方程。划掉既未给出也不需要求的变量，然后使用省略该变量的方程。
Graph areas: For displacement from a v–t graph, areas below the axis are negative. For distance, treat all areas as positive.
图像面积：从 v–t 图求位移时，轴下的面积为负。求路程时，所有面积视为正。
Calculus constants: Always find the constant of integration using the given initial conditions. A missing constant is a frequent mistake.
积分常数：务必用给定的初始条件求出积分常数。遗漏常数是常见错误。

Practice with past WJEC papers, and when stuck, draw a quick sketch of the velocity or displacement function to visualise the motion.

用 WJEC 历年真题练习，当卡住时，快速画出速度或位移函数草图，以直观理解运动。

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A-Level WJEC Maths: Kinematics Key Points Explained | A-Level WJEC 数学：运动学考点精讲