📚 Motion of a Particle in a Straight Line | 直线上的质点运动
Understanding the motion of a particle moving along a straight line is a fundamental topic in IGCSE Mathematics (Edexcel). It introduces the language of kinematics – displacement, velocity, acceleration – and shows how to describe and predict motion using equations, graphs and vector concepts.
理解沿直线运动的质点是 IGCSE 数学（Edexcel）中的一个基础课题。它引入了运动学的语言——位移、速度、加速度——并展示如何利用方程、图形和矢量概念来描述和预测运动。
1. Key Concepts: Scalars and Vectors | 关键概念：标量与矢量
In kinematics, quantities are classified as scalar or vector. A scalar has magnitude only (e.g. speed, distance), while a vector has both magnitude and direction (e.g. velocity, displacement). Direction matters when analysing straight-line motion.
在运动学中，物理量分为标量和矢量。标量只有大小（例如速率、路程），而矢量既有大小又有方向（例如速度、位移）。在分析直线运动时，方向至关重要。
We often choose a positive direction (e.g. 'to the right' or 'upwards') and represent vectors with signs: positive for one direction, negative for the opposite.
我们通常选择一个正方向（例如 '向右' 或 '向上'），并用正负号表示矢量：正号表示一个方向，负号表示相反方向。
2. Displacement, Velocity and Acceleration | 位移、速度和加速度
Displacement (s) is the vector measure of how far an object is from a reference point in a specific direction. Velocity (v) is the rate of change of displacement, and acceleration (a) is the rate of change of velocity.
位移（s）是物体相对于参考点在特定方向上的矢量距离量度。速度（v）是位移的变化率，加速度（a）是速度的变化率。
= total displacement / total time. Average acceleration = change in velocity / time taken.
平均速度 = 总位移 / 总时间。平均加速度 = 速度变化量 / 所用时间。
3. Distance vs Displacement and Speed vs Velocity | 路程与位移、速率与速度的区别
Distance is the total length of the path travelled, a scalar. Displacement is the straight-line distance from start to end with direction. For example, if a particle moves 3 m right then 4 m left, the total distance is 7 m, but the displacement is 1 m left.
路程是运动路径的总长度，是标量。位移是从起点到终点的直线距离并带有方向。例如，如果一个质点先向右移动 3 m，再向左移动 4 m，总路程是 7 m，但位移是向左 1 m。
Similarly, speed is a scalar (rate of distance), while velocity is a vector. Instantaneous speed is the magnitude of instantaneous velocity.
类似地，速率是标量（路程的变化率），而速度是矢量。瞬时速率是瞬时速度的大小。
4. Uniform Acceleration and the SUVAT Equations | 匀加速运动与SUVAT方程
When a particle moves with constant (uniform) acceleration along a straight line, its motion can be described by five quantities: s (displacement), u (initial velocity), v (final velocity), a (acceleration) and t (time).
当质点沿直线以恒定（匀）加速度运动时，其运动可用五个量来描述：s（位移）、u（初速度）、v（末速度）、a（加速度）和 t（时间）。
v = u + a t
Equation 1 links final velocity, initial velocity, acceleration and time.
方程 1 关联末速度、初速度、加速度和时间。
s = u t + ½ a t²
Equation 2 gives displacement when initial velocity, acceleration and time are known.
方程 2 给出了已知初速度、加速度和时间时的位移。
v² = u² + 2 a s
Equation 3 connects final speed, initial speed, acceleration and displacement, without involving time.
方程 3 将末速度、初速度、加速度和位移联系起来，不涉及时间。
s = ½ (u + v) t
Equation 4 expresses displacement as average velocity multiplied by time.
方程 4 用平均速度乘以时间表示位移。
5. Derivation and Use of SUVAT Equations | SUVAT方程的推导与应用
The SUVAT equations can be derived from the definitions of velocity and acceleration when acceleration is constant.
当加速度恒定时，这些SUVAT方程可以从速度和加速度的定义推导出来。
a = (v − u) / t
From the definition of acceleration, rearranging gives v = u + a t, which is Equation 1.
由加速度的定义，整理后得到 v = u + a t，即方程 1。
t under constant acceleration, average velocity = (u + v)/2, and multiplying by time gives s = ½ (u + v) t, Equation 4.
利用匀加速运动中平均速度 = (u + v)/2 这一事实，乘以时间得到 s = ½ (u + v) t，方程 4。
Substituting Equation 1 into Equation 4 yields s = u t + ½ a t², Equation 2. Eliminating t from Equation 1 and Equation 2 results in v² = u² + 2 a s, Equation 3.
将方程 1 代入方程 4 得到 s = u t + ½ a t²，方程 2。从方程 1 和方程 2 中消去 t 得到 v² = u² + 2 a s，方程 3。
When solving problems, identify the known quantities and the one you need to find, then choose the SUVAT equation that contains these four quantities. Watch for missing variable to select efficiently.
解题时，先识别已知量和需要求出的量，然后选择包含这四个量的SUVAT方程。留意缺失的变量可以高效地选择方程。
6. Displacement-Time Graphs | 位移-时间图
A displacement-time (s-t) graph shows how displacement changes with time. For straight-line motion, the slope (gradient) of the graph at any point gives the velocity of the particle.
位移-时间（s-t）图显示位移如何随时间变化。对于直线运动，图中任一点的斜率（梯度）给出质点的速度。
If the graph is a straight line, the velocity is constant. A horizontal line means the particle is stationary.
如果图形是一条直线，则速度恒定。水平线表示质点静止。
If the graph curves, the velocity is changing. The instantaneous velocity is the tangent gradient at that point.
如果图形弯曲，则速度在变化。瞬时速度是曲线上该点的切线斜率。
7. Velocity-Time Graphs | 速度-时间图
A velocity-time (v-t) graph is extremely useful. The gradient gives the acceleration, and the area under the graph gives the displacement.
速度-时间（v-t）图非常有用。斜率给出加速度，图形下的面积给出位移。
For constant acceleration, the v-t graph is a straight sloping line. The area can be calculated using geometry (often a trapezium).
对于匀加速度，v-t 图是一条倾斜直线。面积可以用几何学（通常是梯形）来计算。
If the graph crosses the time axis, velocity changes direction. The area above and below the axis must be treated separately for displacement, but for total distance, take absolute areas.
如果图形穿过时间轴，速度改变方向。位移需要区分轴上方和下方的面积，而总路程则取绝对值之和。
8. Acceleration-Time Graphs | 加速度-时间图
An acceleration-time (a-t) graph mainly highlights how acceleration varies. For uniform acceleration, the graph is a horizontal line. The area under an a-t graph gives the change in velocity.
加速度-时间（a-t）图主要展示加速度如何变化。对于匀加速度，图形是一条水平线。a-t 图下的面积给出速度的变化量。
In most IGCSE problems, acceleration is constant, so graphs are simple. However, recognising the relationship between graphs is important.
在大多数 IGCSE 问题中，加速度是恒定的，因此图形很简单。然而，认识图形之间的关系很重要。
9. Motion Under Gravity | 重力作用下的运动
fine a positive direction (usually up or down) and apply the SUVAT equations with a = ± g.
当质点在重力作用下竖直运动时，加速度恒定，为 g（9.8 m/s²），方向向下。关键是要定义正方向（通常是向上或向下），并用 a = ± g 应用SUVAT方程。
st point, v = 0. Displacement, time of flight and maximum height can be found using SUVAT.
对于以初速度 u 向上抛出的物体，在最高点 v = 0。可以使用SUVAT方程求出位移、飞行时间和最大高度。
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