📚 Displacement along an axis | 沿轴的位移
In one-dimensional kinematics, understanding how an object’s position changes along a straight line is fundamental. Displacement along an axis provides a clear measure of the net change in position, distinguishing it from the total path length travelled. Whether you are analysing the motion of a particle from a velocity function or interpreting a velocity–time graph, mastering displacement is essential for IB Mathematics, especially in the topic of calculus and kinematics.
在一维运动学中，理解物体沿直线的位置变化是基础。沿轴的位移清晰地度量了位置的净变化，将其与走过的总路径长度区分开来。无论你是在通过速度函数分析粒子运动，还是解读速度–时间图，掌握位移对 IB 数学至关重要，特别是在微积分与运动学部分。
1. The Number Line Model | 数轴模型
n arbitrarily for convenience. Positive and negative directions are defined, allowing us to describe both the magnitude and direction of movement.
一维运动使用坐标轴（通常是 x 轴）建模。物体在任意时刻 t 的位置由其位置坐标 x(t) 给出。原点（x = 0）可任意选定以方便计算。定义了正方向和负方向，使我们能够描述运动的量值和方向。
2. Defining Position, Distance and Displacement | 定义位置、路程和位移
It is vital to distinguish three related concepts in kinematics:
区分运动学中三个相关的概念至关重要：
- Position: the coordinate x(t) on the axis, which can be positive, negative, or zero relative to the origin.
- Distance: the total length of the path travelled, always a non-negative scalar.
- Displacement in : the net change in position, defined as Δx = x(final) − x(initial). It can be positive, negative, or zero.
- 位置：轴上的坐标 x(t)，相对原点可正、可负或为零。
- 路程：运动轨迹的总长度，始终为非负标量。
- 位移：位置的净变化，定义为 Δx = x(末) − x(初)。可正、可负或为零。
3. Displacement as the Difference of Positions | 位移作为位置之差
. This simple difference captures the overall effect of the motion, regardless of the path taken between the two points.
若质点从时刻 t₁ 的位置 x₁ 运动到时刻 t₂ 的位置 x₂，其位移 Δx 为 Δx = x₂ − x₁。这个简单的差值反映了运动的总体效果，与两点之间的具体路径无关。
Δx = x(final) − x(initial)
4. Sign Convention and Direction | 符号约定与方向
In one dimension, the sign of displacement indicates direction. A positive displacement means the final position is to the right of the initial position (i.e., in the positive direction), whereas a negative displacement means it is to the left (negative direction). A displacement of zero implies the object returned to its starting point.
在一维中，位移的符号表示方向。正位移意味着末位置在初位置的右侧（即正方向），负位移意味着在左侧（负方向）。位移为零表示物体回到了出发点。
5. Displacement from a Velocity–Time Graph | 从速度–时间图中求位移
The area between the velocity–time graph and the time axis represents displacement over that interval. Areas above the time axis contribute positive displacement, while areas below contribute negative displacement. The net displacement is the algebraic sum of these signed areas.
速度–时间图与时间轴之间的面积代表该时段内的位移。时间轴上方的面积贡献正位移，下方的面积贡献负位移。净位移就是这些有符号面积的代数和。
6. Calculating Displacement Using Integration | 用积分计算位移
When velocity is expressed as a function of time v(t), displacement s over the interval [a, b] is the definite integral of velocity. This is a direct consequence of the fundamental theorem of calculus, linking differentiation and integration of the position function.
当速度表示为时间的函数 v(t) 时，在区间 [a, b] 上的位移 s 就是速度的定积分。这是微积分基本定理的直接结果，将位置函数的微分和积分联系了起来。
s = ∫ₐᵇ v(t) dt
7. Relationship Between Displacement and Total Distance | 位移与总路程的关系
Total distance travelled is the integral of the absolute value of velocity, or equivalently the sum of the absolute areas bounded by the v–t graph. While displacement can be zero for a round trip, total distance is always positive and accumulates all path increments regardless of direction.
总路程是速度绝对值的积分，等价于速度–时间图所围各块面积的绝对值之和。一次往返的位移可能为零，但总路程始终为正，并累积所有方向的路径增量。
Total distance = ∫ₐᵇ |v(t)| dt
8. Displacement in Uniformly Accelerated Motion (SUVAT) | 匀加速运动中的位移 (SUVAT)
For motion with constant acceleration a, initial velocity u, the displacement s in time t can be found using the SUVAT equations. These relationships are derived from the definitions of velocity and acceleration and are extremely useful in IB kinematics problems.
对于恒定加速度 a 和初速度 u 的运动，时间 t 内的位移 s 可用 SUVAT 方程求出。这些关系源自速度和加速度的定义，在 IB 运动学问题中极为有用。
s = ut + ½at² or s = (u + v)t / 2, where v = u + at
9. Worked Example: Constant Velocity | 例题：匀速运动
t x = 2 m. Find its displacement between t = 0 s and t = 4 s, and its position at t = 4 s.
质点沿 x 轴以恒定速度 v = 5 m/s 运动。t = 0 s 时位于 x = 2 m。求 t = 0 s 到 t = 4 s 的位移及 t = 4 s 时的位置。
Solution: = v × Δt = 5 × 4 = 20 m. Position x(4) = initial position + displacement = 2 + 20 = 22 m.
解：位移 = v × Δt = 5 × 4 = 20 m。位置 x(4) = 初位置 + 位移 = 2 + 20 = 22 m。
10. Worked Example: Variable Velocity | 例题：变速运动
moves with velocity v(t) = 3t² − 4t (m/s). Find the displacement from t = 0 s to t = 3 s.
质点运动速度为 v(t) = 3t² − 4t (m/s)。求 t = 0 s 到 t = 3 s 的位移。
Solution: s = ∫₀³ (3t² − 4t) dt = [t³ − 2t²]₀³ = (27 − 18) − (0) = 9 m. The net displacement is 9 m.
解：s = ∫₀³ (3t² − 4t) dt = [t³ − 2t²]₀³ = (27 − 18) − (0) = 9 m。净位移为 9 m。
11. Common Pitfalls and Exam Tips | 常见错误与考试技巧
Always check whether a question asks for displacement or total distance. When reading a velocity function, pay attention to intervals where velocity becomes zero or changes sign; these affect the area calculations. In integration, ensure limits are correctly applied and remember that displacement is the integral of velocity, not speed. Practise switching between graphical, analytical, and SUVAT approaches to reinforce your understanding.
务必分清题目是求位移还是总路程。分析速度函数时，注意速度为零或变号的区间，这会影响面积计算。积分时确保上下限正确，记住位移是速度的积分而不是速率。多练习图解、解析和 SUVAT 方程之间的转换，以加深理解。
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