📚 Using Integration to Find Displacement and Total Distance | 利用积分法求解位移与路程
In one-dimensional kinematics, the velocity function v(t) is the rate of change of displacement s(t). By the Fundamental Theorem of Calculus, integrating v(t) over a time interval gives the net change in position. However, to find the total distance travelled, we must integrate the absolute value of velocity — the speed. This distinction is essential in A-level Mechanics and is a frequent source of lost marks.
在一维运动学中，速度函数 v(t) 是位移 s(t) 的变化率。根据微积分基本定理，对 v(t) 在时间区间上积分得到位置的净变化。然而，若要计算总路程，必须对速度的绝对值——即速率——进行积分。这一区别在 A-Level 力学中至关重要，同时也是常见的失分点。
1. Displacement vs Distance | 位移与路程的区别
Displacement is a vector quantity that measures the straight-line change in position from the start point to the end point. It can be positive, negative or zero.
位移是矢量，衡量从起点到终点的直线位置变化，可以为正、负或零。
Total distance is a scalar quantity that measures the length of the actual path travelled. It is always non-negative.
总路程是标量，衡量实际经过路径的长度，始终为非负数。
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Displacement answers ‘how far from the start in a given direction’; distance answers ‘how much ground was covered’.
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位移回答“在给定方向上离起点有多远”；路程回答“总共走了多少路”。
2. Integrating Velocity to Obtain Displacement | 对速度积分得到位移
For a particle moving along a straight line with velocity v(t), the displacement between times t₁ and t₂ is given by:
对于沿直线运动、速度为 v(t) 的质点，在时刻 t₁ 与 t₂ 之间的位移为：
s(t₂) – s(t₁) = ∫t₁t₂ v(t) dt
This integral accumulates every signed change in position. A negative velocity subtracts from the total.
该积分累加每一个带符号的位置变化。速度为负时会使结果减少。
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If v(t) > 0 throughout, the particle moves in the positive direction and displacement equals distance.
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如果整个过程中 v(t) > 0，质点向正方向运动，位移等于路程。
3. Integrating Speed to Obtain Total Distance | 对速率积分得到总路程
Total distance is obtained by integrating the speed, which is |v(t)|:
总路程通过对速率 |v(t)| 进行积分得到：
Total distance = ∫t₁t₂ |v(t)| dt
Using the absolute value ensures that every piece of motion, whether forward or backward, contributes positively.
使用绝对值可以确保无论向前还是向后的每一段运动都正向累加。
4. Why the Absolute Value Matters | 为什么绝对值很重要
When a particle reverses direction, its velocity changes sign. For example, a ball thrown upwards has positive velocity while rising and negative velocity while falling. Simply integrating v(t) cancels the upward and downward contributions.
当质点反向运动时，速度发生变号。例如，竖直上抛的小球上升时速度为正，下落时速度为负。如果直接对 v(t) 积分，上升和下降的贡献会相互抵消。
Distance, however, must count the upward and downward journeys separately. This is why we integrate |v(t)|, not v(t).
然而，路程必须分别计算上升段和下降段。这就是为什么我们要对 |v(t)| 而非 v(t) 积分。
5. Worked Example 1: Constant Direction Motion | 例题 1：单向直线运动
A particle moves with velocity v(t) = 2t m/s for 0 ≤ t ≤ 3. Find the displacement and the total distance.
质点以速度 v(t) = 2t m/s 在 0 ≤ t ≤ 3 内运动，求位移和总路程。
Since v(t) ≥ 0 on this interval, displacement and distance coincide.
由于区间上 v(t) ≥ 0，位移和路程相同。
Displacement = ∫₀³ 2t dt = [t²]₀³ = 9 m
Total distance = ∫₀³ |2t| dt = 9 m
Because the velocity never becomes negative, no further splitting is required.
因为速度从未变负，无需再分段处理。
6. Worked Example 2: Reversing Direction | 例题 2：反向运动
Consider v(t) = 3t² – 12t + 9 m/s, for 0 ≤ t ≤ 4. Find the displacement and total distance.
设 v(t) = 3t² – 12t + 9 m/s，0 ≤ t ≤ 4，求位移和总路程。
Factorising: v(t) = 3(t – 1)(t – 3). The velocity is zero at t = 1 and t = 3, so we split the interval at these points.
因式分解得 v(t) = 3(t – 1)(t – 3)。速度在 t = 1 和 t = 3 处为零，因此在这两点分割区间。
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For 0 < t < 1, v(t) > 0.
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当 0 < t < 1 时，v(t) > 0。
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For 1 < t < 3, v(t) < 0.
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当 1 < t < 3 时，v(t) < 0。
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For 3 < t < 4, v(t) > 0.
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当 3 < t < 4 时，v(t) > 0。
The displacement is the single integral of v(t):
位移是对 v(t) 的一次积分：
s(4) – s(0) = ∫₀⁴ (3t² – 12t + 9) dt = [t³ – 6t² + 9t]₀⁴ = 4 m
The total distance splits the integral wherever v changes sign:
总路程在速度变号处分段积分：
Distance = ∫₀¹ v dt – ∫₁³ v dt + ∫₃⁴ v dt
Evaluating each piece:
计算每一段：
∫₀¹ v dt = [t³ – 6t² + 9t]₀¹ = 4
∫₁³ v dt = [t³ – 6t² + 9
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