📚 Kinematics: Motion of a Particle in a Straight Line | 运动学：质点沿直线运动
In the mechanics section of Edexcel IGCSE Mathematics, we often model a moving object as a particle — a point mass with no size — and study its motion along a straight line. This topic combines algebraic manipulation with graph interpretation, and appears regularly in the exam through structured questions on distance–time graphs, speed–time graphs, and the equations of constant acceleration.
在 Edexcel IGCSE 数学的力学部分中，我们经常把运动的物体建模为一个质点——一个没有大小的点——并研究它沿直线的运动。这一主题将代数运算与图像解读结合起来，在考试中经常通过距离—时间图像、速度—时间图像以及匀变速直线运动公式的结构化问题出现。
1. Position, Displacement and Distance | 位置、位移与路程
The position a particle is its coordinate on a chosen axis, for example x = 3 m from the origin O. The displacement is the change in position, written as Δx; it has both magnitude and direction, so it can be positive or negative. The distance travelled, however, is the total length of the path covered, and it is always positive or zero.
位置是质点在所选坐标轴上的坐标，例如离原点 O 为 x = 3 m。位移是位置的变化量，记作 Δx；它既有大小又有方向，因此可以为正也可以为负。而路程是物体运动轨迹的总长度，始终为正或零。
it finishes 1 m to the left of the start. The total distance travelled is 3 m + 4 m = 7 m.
设想一个质点从 x = 2 m 运动到 x = 5 m，再返回 x = 1 m。它相对起点的位移是 1 − 2 = −1 m，表示终点在起点左侧 1 m 处。而它走过的总路程是 3 m + 4 m = 7 m。
- Displacement is a vector: sign tells you the direction.
- 位移是矢量：正负号表示方向。
- Distance is a scalar: always positive or zero.
- 路程是标量：始终为正或零。
- Units for both: metres (m).
- 两者的单位都是米（m）。
2. Speed and Velocity | 速率与速度
Speed is the distance travelled per unit time; it is a scalar, so its value is never negative. Velocity is the displacement per unit time; it is a vector, so it carries a sign that indicates the direction of motion.
速率是单位时间内通过的路程；它是标量，因此永远不会为负。速度是单位时间内的位移；它是矢量，带有表示运动方向的正负号。
For uniform motion along a straight line, we use:
对于沿直线的匀速运动，我们使用：
speed = distance ÷ time
velocity = displacement ÷ time
nd (m/s) or kilometres per hour (km/h). To convert: 1 m/s = 3.6 km/h, so 36 km/h = 10 m/s.
在 IGCSE 试题中，速率和速度的单位是米每秒（m/s）或千米每小时（km/h）。换算关系：1 m/s = 3.6 km/h，因此 36 km/h = 10 m/s。
If a particle returns to its starting point, its average velocity is zero, but its average speed is greater than zero. This distinction is a favourite test point in exams.
如果质点回到起点，则其平均速度为零，但平均速率大于零。这个区分是考试中最受青睐的考点之一。
3. Acceleration | 加速度
Acceleration n object slows down, its acceleration is opposite to its velocity; we then call it deceleration or retardation.
加速度是速度的变化率，定义为 a = Δv ÷ Δt。它是一个矢量。当物体减速时，加速度方向与速度方向相反，我们称之为减速度。
For uniform acceleration along a straight line, the acceleration is constant over time. For example, if a particle's velocity increases from 4 m/s to 10 m/s in 3 s, then:
对于沿直线的匀变速运动，加速度在时间上是恒定的。例如，如果质点的速度在 3 s 内从 4 m/s 增加到 10 m/s，则：
a = (10 − 4) ÷ 3 = 2 m/s²
- Positive acceleration: velocity is increasing in the chosen positive direction.
- 加速度为正：速度沿所选正方向增大。
- Negative acceleration: velocity is decreasing in that direction.
- 加速度为负：速度沿该方向减小。
- Unit: metre per second squared (m/s²).
- 单位：米每二次方秒（m/s²）。
4. Displacement–Time Graphs | 位移—时间图像
A displacement–time graph shows the position of a particle against time. Its gradient gives the velocity of the particle.
位移—时间图像显示质点位置随时间的变化。它的斜率给出质点的速度。
- Gradient positive → moving in the positive direction.
- 斜率为正 → 沿正方向运动。
- Gradient negative → moving in the negative direction.
- 斜率为负 → 沿负方向运动。
- Gradient zero (horizontal line) → particle is at rest.
- 斜率为零（水平线）→ 质点静止。
- Steeper gradient → greater speed.
- 斜率越陡 → 速率越大。
- A curved line → velocity is changing, so acceleration is not zero.
- 曲线 → 速度在变化，因此加速度不为零。
m in 20 s at constant speed, the graph is a straight line with gradient 100 ÷ 20 = 5 m/s.
例如，如果质点以恒定速度在 20 s 内从 x = 0 运动到 x = 100 m，图像是一条直线，其斜率为 100 ÷ 20 = 5 m/s。
5. Velocity–Time Graphs | 速度—时间图像
A velocity–time graph shows how velocity changes with time. Two pieces of information are crucial here.
速度—时间图像显示速度随时间的变化。这里有两个关键信息。
- The gradient of the graph gives the acceleration.
- 图像的斜率给出加速度。
- The area between the graph and the time axis gives the displacement.
- 图像与时间轴之间的面积给出位移。
If the velocity is constant, the graph is a horizontal line and the acceleration is zero. If the line slopes up, the particle is accelerating; if it slopes down, the particle is decelerating.
如果速度恒定，图像是水平线，加速度为零。如果线段向上倾斜，质点在加速；如果线段向下倾斜，质点在减速。
Remember that the area below the time axis is negative displacement. To find the total distance travelled, add the absolute values of the areas on both sides of the axis. To find the final displacement, add the signed areas.
注意，时间轴下方的面积为负位移。要计算总路程，应把轴两侧面积的绝对值相加。要计算最终位移，则把带符号的面积相加。
For example, a triangle of base 10 s and height 20 m/s has area ½ × 10 × 20 = 100 m — the displacement during that interval.
例如，底为 10 s、高为 20 m/s 的三角形，面积为 ½ × 10 × 20 = 100 m——即该时间段内的位移。
6. The SUVAT Equations | SUVAT 运动学公式
When acceleration is constant, the five quantities below are linked by four standard equations:
当加速度恒定时，下面五个量由四个标准公式联系起来：
- s = displacement (m)
- s = 位移（m）
- u = initial velocity (m/s)
- u = 初速度（m/s）
- v = final velocity (m/s)
- v = 末速度（m/s）
- a = constant acceleration (m/s²)
- a = 恒定加速度（m/s²）
- t = time taken (s)
- t = 所用时间（s）
v = u + at
s = ½(u + v)t
s = ut + ½at²
v² = u² + 2as
Each equation omits one variable, so choose the equation that contains exactly the quantities you know and the one you need to find.
每个公式都缺少一个变量，因此选择恰好包含已知量和待求量的公式即可。
| Equation | 公式 | Omits | 不含 | Best used when | 适用情形 |
|---|---|---|
| v = u + at | s | time is given, displacement is not needed | 给时间且无需位移 |
| s = ½(u + v)t | a | acceleration is unknown | 加速度未知 |
| s = ut + ½at² | v | final velocity is unknown | 末速度未知 |
| v² = u² + 2as | t | time is not given | 未给出时间 |
These equations are only valid when acceleration is constant. If acceleration changes, you must use graphs or split the motion into stages.
这些公式仅在加速度恒定时成立。如果加速度变化，则必须使用图像或将运动分成多个阶段。
7. Worked Example — Constant Acceleration | 例题：匀变速直线运动
A particle passes point A with velocity 6 m/s and accelerates uniformly at 2 m/s² for 5 s. Find:
一个质点以 6 m/s 的速度经过 A 点，并以 2 m/s² 的加速度匀加速运动 5 s。求：
(a) the velocity after 5 s;
（a）5 s 后的速度；
Using v = u + at:
使用 v = u + at：
v = 6 + 2 × 5 = 16 m/s
(b) the displacement from A in this time;
（b）这段时间内离开 A 点的位移；
Using s = ut + ½at²:
使用 s = ut + ½at²：
s = 6 × 5 + ½ × 2 × 5² = 30 + 25 = 55 m
(c) the velocity of
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图