📚 Constant Acceleration Formulae 2 | 匀加速运动公式（二）
In this article, we continue our study of constant acceleration formulae, focusing on the less obvious applications: using the formulae without initial velocity, solving problems with two or more objects, and combining algebraic manipulation with physical reasoning. These skills are essential for Edexcel A-Level Mathematics, particularly in mechanics sections of both pure and applied papers.
本文继续深入探讨匀加速运动公式，重点放在不太直观的应用：不含初速度的公式用法、两个或多个物体的运动问题，以及代数运算与物理推理的结合。这些技能是 Edexcel A-Level 数学考试的核心内容，尤其在力学部分中经常出现。
1. The Four Key Formulae Revisited | 回顾四条重要公式
The four constant acceleration formulae are: v = u + at, s = ut + ½at², s = ½(u + v)t, and v² = u² + 2as. Here u is initial velocity, v is final velocity, a is constant acceleration, t is time, and s is displacement.
四条匀加速公式分别是：v = u + at，s = ut + ½at²，s = ½(u + v)t，v² = u² + 2as。其中 u 为初速度，v 为末速度，a 为恒定加速度，t 为时间，s 为位移。
Each formula involves four of the five variables, so when any three are known, the remaining two can be found. Choosing the correct formula is often the key to solving a problem quickly.
每条公式涉及五个变量中的四个，因此当已知任意三个变量时，可以求出另外两个。正确选择公式往往是快速解题的关键。
2. Working Without Initial Velocity | 初速度未知时的处理
Sometimes the initial velocity u is not given, but we can still use the formulae. For example, a particle is observed to travel 20 m in the first 3 seconds and 35 m in the next 3 seconds, with constant acceleration. We can write two equations using s = ut + ½at², then solve simultaneously.
有时初速度 u 并未直接给出，但我们依然可以使用公式。例如，一个质点在最初 3 秒内运动 20 米，接下来的 3 秒内运动 35 米，且加速度恒定。我们可以利用 s = ut + ½at² 列出两个方程，然后联立求解。
For the first interval: 20 = 3u + ½a(3²). For the second interval, the displacement over the first 6 seconds is 20 + 35 = 55, so 55 = 6u + ½a(6²). Solving gives u = 4 m/s and a = 16/9 m/s².
对于第一个时间段：20 = 3u + ½a(3²)。对于第二个时间段，前 6 秒的总位移为 20 + 35 = 55，因此 55 = 6u + ½a(6²)。联立解得 u = 4 m/s，a = 16/9 m/s²。
This technique is powerful because it avoids the need to know the velocity at any instant. Always remember that displacement is cumulative, so the second interval displacement equals s(total to 6s) minus s(total to 3s).
这种技巧非常强大，因为它不需要知道任何瞬时的速度。务必记住位移是累积的，因此第二个时间段的位移等于前 6 秒的总位移减去前 3 秒的总位移。
3. Using v² = u² + 2as When Time Is Not Needed | 无需时间时使用 v² = u² + 2as
When a problem does not involve time, the formula v² = u² + 2as is often the most direct. For example, a car accelerates uniformly from 10 m/s to 30 m/s over a distance of 200 m. Find the acceleration.
当问题中不涉及时间时，公式 v² = u² + 2as 往往最为直接。例如，一辆汽车从 10 m/s 匀加速到 30 m/s，行驶距离为 200 米，求加速度。
Substitute: 30² = 10² + 2a(200), so 900 = 100 + 400a, giving a = 2 m/s².
代入：30² = 10² + 2a(200)，即 900 = 100 + 400a，解得 a = 2 m/s²。
This formula is particularly useful for vertical motion under gravity, where the acceleration is g = 9.8 m/s² and the time is often not given.
该公式特别适用于重力作用下的竖直运动，其中加速度为 g = 9.8 m/s²，且时间通常不给出。
4. Sign Conventions: The Most Common Pitfall | 正负号约定：最常见的陷阱
All vector quantities in these formulae must be assigned a consistent positive direction. If upward is positive, then a = −9.8 m/s² for an object moving under gravity. Mixing signs is the most common source of errors.
公式中的所有矢量量都必须指定一致的正方向。如果取向上为正，则在重力作用下的物体 a = −9.8 m/s²。混淆正负号是最常见的错误来源。
Example: A ball is thrown upward with speed 15 m/s. Find the time to reach the highest point. Taking upward as positive, v = 0, u = 15, a = −9.8. Using v = u + at gives 0 = 15 − 9.8t, so t ≈ 1.53 s.
例：一个小球以 15 m/s 的初速度竖直上抛。求到达最高点的时间。取向上为正，v = 0，u = 15，a = −9.8。利用 v = u + at 得 0 = 15 − 9.8t，因此 t ≈ 1.53 秒。
Always state your positive direction before writing any equation. This small habit saves many marks in the exam.
在写出任何方程之前，务必说明正方向。这个小习惯可以在考试中为你节省很多分数。
5. Solving Simultaneous Equations with Constant Acceleration | 匀加速运动中的联立方程求解
When two unknown quantities appear, you need two independent equations. Typical pairs are u and a, or s and t, or a and t. Use the formula that best matches the given information.
当出现两个未知量时，你需要两个独立的方程。典型的未知量组合是 u 和 a，或 s 和 t，或 a 和 t。根据已知信息选择最合适的公式。
Example: A particle passes point A with speed u and accelerates uniformly. It reaches point B, 72 m away, after 6 seconds with speed 18 m/s. Find u and a.
例：一质点以速度 u 通过 A 点并匀加速。经过 6 秒到达距 A 72 米的 B 点，速度为 18 m/s。求 u 和 a。
Use s = ½(u + v)t: 72 = ½(u + 18) × 6, so u + 18 = 24, giving u = 6 m/s. Then use v = u + at: 18 = 6 + 6a, so a = 2 m/s².
利用 s = ½(u + v)t：72 = ½(u + 18) × 6，得 u + 18 = 24，即 u = 6 m/s。再利用 v = u + at：18 = 6 + 6a，得 a = 2 m/s²。
Write down the known variables first: u = ?, v = 18, a = ?, s = 72, t = 6. Then the choice of formula becomes obvious.
先写出已知变量：u = ?，v = 18，a = ?，s = 72，t = 6。这样公式的选择就非常明了。
6. Two-Particle Problems: Meeting Points | 双质点问题：相遇点
When two objects move independently, we often set up separate equations for each, then equate either their displacements or their times at the meeting moment.
当两个物体独立运动时，我们通常分别为每个物体建立方程，然后在相遇时刻令它们的位移或时间相等。
Example: Car A starts from rest at O with acceleration 2 m/s². Two seconds later, car B passes O at constant speed 15 m/s. When and where do they meet?
例：汽车 A 从 O 点由静止出发，加速度为 2 m/s²。2 秒后，汽车 B 以 15 m/s 的恒定速度经过 O 点。问何时、何地两车相遇？
Let t be the time after A starts. For A: s = ½(2)t² = t². For B: it has been moving for (t − 2) seconds, so s = 15(t − 2). Equating: t² = 15(t − 2), so t² − 15t + 30 = 0. Solving gives t ≈ 2.35 s or 12.65 s. The first root is before B has covered any positive distance? Check: t = 2.35 gives s ≈ 5.5 m, B distance = 15(0.35) = 5.25 m, close due to rounding. Actually exact solving gives t = (15 ± √105)/2. The meaningful meeting after B starts is t ≈ 12.65 s, s ≈ 160 m.
设 t 为 A 出发后的时间。对于 A：s = ½(2)t² = t²。对于 B：它已经运动了 (t − 2) 秒，所以 s = 15(t − 2)。令两者相等：t² = 15(t − 2)，即 t² − 15t + 30 = 0，解得 t ≈ 2.35 秒或 12.65 秒。经检验，t ≈ 12.65 秒是 B 出发后两车真正相遇的时刻，对应位移 s ≈ 160 米。
Be careful with time shifts: if B starts later, its time variable is the original t minus the delay. Always define the origin of time clearly.
注意时间错位：如果 B 后出发，它的时间变量就是原来的 t 减去延迟量。一定要明确时间零点的定义。
7. Vertical Motion Under Gravity | 重力作用下的竖直运动
For vertical motion, the acceleration is constant at g = 9.8 m/s² downward. We choose a positive direction, usually upward, then apply the same four formulae, taking a = −9.8 if upward is positive.
对于竖直运动，加速度恒为 g = 9.8 m/s² 向下。我们通常取向上为正方向，然后应用相同的四条公式，如果向上为正则 a = −9.8。
Example: A stone is thrown vertically upward with speed 12 m/s from the top of a cliff 25 m high. Find the velocity with which it hits the sea below.
例：一块石头从 25 米高的悬崖顶部以 12 m/s 的初速度竖直上抛。求石头落到下方海面时的速度。
Take upward as positive. The displacement to the sea is s = −25 m. u = 12 m/s, a = −9.8. Use v² = u² + 2as: v² = 12² + 2(−9.8)(−25) = 144 + 490 = 634, so v = ±√634. Since the stone is moving downward, v = −√634 ≈ −25.2 m/s. The speed is 25.2 m/s.
取向上为正。落到海面的位移 s = −25 m。u = 12 m/s，a = −9.8。利用 v² = u² + 2as：v² = 12² + 2(−9.8)(−25) = 144 + 490 = 634，所以 v = ±√634。由于石头向下运动，v = −√634 ≈ −25.2 m/s，速率为 25.2 m/s。
Note that we did not need to split the motion into upward and downward phases. The formula handles the whole motion in one go because acceleration is constant.
注意，我们不需要将运动分为上升和下降两个阶段。由于加速度恒定，公式可以一次性处理整个运动过程。
8. Variable Acceleration vs Constant Acceleration | 变加速度与匀加速度的区别
The formulae in this chapter are only valid when acceleration is constant. If acceleration depends on time or position, you must use calculus: v = ∫a dt, s = ∫v dt, and a = v dv/ds.
本章中的公式仅在加速度恒定的时候成立。如果加速度随时间或位置变化，你必须使用微积分：v = ∫a dt，s = ∫v dt，以及 a = v dv/ds。
In Edexcel exams, the question will often state “constant acceleration” or “uniform acceleration”. If not, look for words like “decelerates uniformly” or “moves with constant force”, which imply constant acceleration.
在 Edexcel 考试中，题目通常会说明”恒加速度”或”匀加速度”。如果没有，请留意”匀减速”或”恒定的力”等表述，这些也暗示加速度恒定。
When solving problems involving maximum height or speed, remember that at the highest point the velocity is zero. This fact provides an extra equation.
在求解最大高度或最大速度的问题时，记住在最高点速度为零。这一事实给出了一个额外的方程。
9. Multi-Stage Motion | 多阶段运动
Many real problems involve two or more stages, each with different constant acceleration. For example, a train accelerates from rest, travels at constant speed, then decelerates to stop. Each stage must be analysed separately, then combined.
许多实际问题包含两个或多个阶段，每个阶段具有不同的恒定加速度。例如，火车从静止加速，匀速行驶，然后减速停止。每个阶段必须单独分析，然后组合起来。
Key quantities such as the final velocity of one stage become the initial velocity of the next. The total displacement is the sum of the displacements of all stages.
关键量如某一阶段的末速度会成为下一阶段的初速度。总位移等于各阶段位移之和。
Example: A particle accelerates from rest at 2 m/s² for 5 seconds, then continues at constant speed for 3 seconds, then decelerates at 4 m/s² until it stops. Find the total distance.
例：一个质点从静止开始以 2 m/s² 加速 5 秒，然后匀速运动 3 秒，再以 4 m/s² 减速至停止。求总距离。
Stage 1: v = 0 + 2(5) = 10 m/s, s₁ = ½(0 + 10)(5) = 25 m. Stage 2: s₂ = 10 × 3 = 30 m. Stage 3: using v² = u² + 2as, 0 = 10² + 2(−4)s₃, so s₃ = 100/8 = 12.5 m. Total = 25 + 30 + 12.5 = 67.5 m.
阶段1：v = 0 + 2(5) = 10 m/s，s₁ = ½(0 + 10)(5) = 25 m。阶段2：s₂ = 10 × 3 = 30 m。阶段3：利用 v² = u² + 2as，0 = 10² + 2(−4)s₃，得 s₃ = 100/8 = 12.5 m。总距离 = 25 + 30 + 12.5 = 67.5 m。
Always draw a velocity-time graph for multi-stage problems; the area under the graph gives displacement, and the slopes give acceleration. This provides a powerful cross-check.
对于多阶段问题，务必画速度-时间图；图线与时间轴围成的面积给出位移，斜率给出加速度。这是一种强有力的检验手段。
10. Using Graphs to Solve Kinematics Problems | 用图像解决运动学问题
A velocity-time graph can often replace algebraic manipulation. For constant acceleration, the graph is a straight line. The gradient is acceleration, the area under the graph between two times is displacement.
速度-时间图通常可以替代代数运算。对于匀加速运动，图像是一条直线。斜率为加速度，图线在时间段内与时间轴围成的面积为位移。
Example: A cyclist accelerates uniformly from 4 m/s to 10 m/s in 12 seconds. Find the distance travelled.
例：一名骑自行车的人从 4 m/s 匀加速到 10 m/s，用时 12 秒。求行驶的距离。
The area of the trapezium is ½(4 + 10) × 12 = 84 m. This matches the formula s = ½(u + v)t.
梯形的面积为 ½(4 + 10) × 12 = 84 m。这与公式 s = ½(u + v)t 一致。
If a particle is thrown upward, the velocity-time graph crosses the time axis at the highest point. The two triangles on either side have equal area if it returns to the same height, meaning equal displacement up and down.
如果质点被竖直上抛，速度-时间图会在最高点处穿过时间轴。如果它回到同一高度，则图线两侧的三角形面积相等，表示上升和下降的位移大小相等。
11. Common Exam Question Types | 常见考试题型
In Edexcel Mechanics, typical constant acceleration questions include: finding stopping distance, catching up problems, projectiles thrown vertically, and calculating reaction times. One classic type involves a vehicle braking: you are given initial speed, deceleration, and asked for thinking distance plus braking distance.
在 Edexcel 力学中，典型的匀加速问题包括：求停车距离、追及问题、竖直抛体，以及反应时间的计算。一个经典类型是车辆刹车：给出初速度和减速度，要求思考距离加上制动距离。
Another type asks for the time when velocity is zero, or displacement after a certain time. Remember that a negative displacement simply means the object is on the opposite side of the starting point along the positive direction.
另一类问题求速度为零的时刻，或某时刻后的位移。记住，负位移仅表示物体位于起点沿正方向的反侧。
Sometimes a question gives a table of data for displacement at different times. You can use finite differences to check if acceleration is constant, then find u and a via simultaneous equations.
有时题目会给出不同时刻位移的数据表。你可以利用有限差分来判断加速度是否恒定，然后通过联立方程求 u 和 a。
12. Worked Example: Full Edexcel-Style Problem | 完整例题：Edexcel 风格题目
Problem: A particle P moves with constant acceleration. Initially P is at the origin and moving with velocity (3i − 2j) m/s. After 4 seconds, the velocity is (5i + 6j) m/s. Find the magnitude of the acceleration and the displacement after 4 seconds.
题目：质点 P 以恒定加速度运动。初始时 P 在原点，速度为 (3i − 2j) m/s。4 秒后速度为 (5i + 6j) m/s。求加速度的大小以及 4 秒后的位移。
Since acceleration is constant, a = (v − u)/t = [(5i + 6j) − (3i − 2j)] / 4 = (2i + 8j) / 4 = (0.5i + 2j) m/s². The magnitude is √(0.5² + 2²) = √4.25 ≈ 2.06 m/s².
因为加速度恒定，a = (v − u)/t = [(5i + 6j) − (3i − 2j)] / 4 = (2i + 8j) / 4 = (0.5i + 2j) m/s²。其大小为 √(0.5² + 2²) = √4.25 ≈ 2.06 m/s²。
Displacement s = ut + ½at² = (3i − 2j)(4) + ½(0.5i + 2j)(4²) = (12i − 8j) + (8i + 16j) = (20i + 8j) m. The magnitude of displacement is √(20² + 8²) = √464 ≈ 21.54 m.
位移 s = ut + ½at² = (3i − 2j)(4) + ½(0.5i + 2j)(4²) = (12i − 8j) + (8i + 16j) = (20i + 8j) m。位移大小为 √(20² + 8²) = √464 ≈ 21.54 m。
This vector approach is common in A-Level questions, especially when combining motion in the x and y directions. Treat each direction independently using the same scalar formulae.
这种向量方法在 A-Level 题目中非常常见，尤其是结合 x 和 y 方向的运动时。每个方向都使用相同的标量公式独立处理。
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