题型概述：一维运动、抛体与变加速度Question type: 1-D motion, projectiles and non-uniform acceleration
Top Gold运动学是 Section 1 的"送分热身区"：每年约 5 分，常以 Q1(a)–(c) 的 3–5 分小题出现（落体、追及、抛体），偶尔有一道 6–7 分的变加速度或抛体综合题。物理本身不难，失分集中在四处：正负号（向上为正时 a = −g）、多段运动当成一段、km/h 没换成 m/s，以及加速度不恒定时还在套 suvat。Kinematics is Section 1's warm-up zone: about 5 marks a year, usually as 3–5 mark items in Q1(a)–(c) (falling objects, chases, projectiles), sometimes with a 6–7 mark non-uniform-acceleration or projectile question. The physics is simple; marks are lost in four places: signs (with up positive, a = −g), treating multi-stage motion as one stage, not converting km/h to m/s, and using suvat when the acceleration is not constant.
1suvat 与运动图像suvat and motion graphs
加速度恒定时，五个量 s、u、v、a、t 由下面的方程联系，知道任意三个就能求另外两个。图像给出同样的信息：v–t 图的斜率 = a，面积 = 位移 s（横轴下方的面积为负）；s–t 图的斜率 = v。When the acceleration is constant, the five quantities s, u, v, a, t are linked by the equations below; any three give the other two. Graphs carry the same information: on a v–t graph the gradient = a and the area = displacement s (area below the axis counts as negative); on an s–t graph the gradient = v.
v = u + at, s = ut + ½at², v² = u² + 2as, s = ½(u + v)t最后一个式子说：匀加速时平均速度就是 (u + v)/2。追及题里"同时出发、同时到达"⇒ 平均速度相等，常能一步解题。The last one says that for uniform acceleration the average velocity is (u + v)/2. In chase problems "same start, same finish time" ⇒ equal average velocities, which often solves the problem in one line.
绿色面积在横轴上方（向正方向走），橙色在下方（往回走）。取 u = 10、a = −4：物体 2.5 s 时掉头，位移和路程开始不同——竖直上抛就是这种情况。Green area lies above the axis (moving forwards), orange below (moving back). With u = 10, a = −4 the object turns round at 2.5 s and displacement and distance start to differ — exactly what happens to a ball thrown upwards.
2分段运动与相遇 / 追及Multi-stage motion, meeting and chasing
分段运动：每换一种加速度就是新的一段，每段单独用 suvat；上一段的末速度就是下一段的初速度（段与段靠速度"接力"）。相遇 / 追及：两个物体"相遇"就是同一时刻、同一位置，把两者的位置写成同一个时钟 t 的函数然后令其相等。晚出发的物体要用 (t − Δt) 作它自己的飞行时间。Multi-stage motion: every change of acceleration starts a new stage; apply suvat to each stage separately, and the final velocity of one stage is the initial velocity of the next (the stages are linked by velocity). Meeting / chasing: two bodies "meet" when they are at the same place at the same time. Write both positions as functions of one clock t and set them equal; a body that starts later uses (t − Δt) as its own time of flight.
sA(t) = sB(t) ⇔ v–t 图上两条线下的面积相等graph: equal areas under the two lines两个物体加速度相同（例如都只受重力）时，t² 项互相抵消，相对速度保持不变——相遇问题退化成"匀速追赶"。If both bodies have the same acceleration (e.g. both only under gravity), the t² terms cancel and the relative velocity stays constant — the meeting problem becomes a constant-speed chase.
3二维抛体与变加速度：分解与积分Projectiles and non-uniform acceleration: resolve and integrate
抛体：忽略空气阻力时水平方向不受力，vx 不变；竖直方向加速度为 −g。两个方向共用同一个 t，各自用 suvat，最后用勾股定理和 arctan 合成速度的大小和方向。变加速度：a 随时间变化时 suvat 失效，要回到定义 a = dv/dt、v = ds/dt，用积分（即 a–t 图、v–t 图下的面积）求 Δv 和 Δs。Projectiles: ignoring air resistance there is no horizontal force, so vx is constant; vertically the acceleration is −g. Both directions share the same t; apply suvat to each, then combine with Pythagoras and arctan for the speed and direction. Non-uniform acceleration: when a changes with time suvat fails; go back to the definitions a = dv/dt and v = ds/dt and integrate (the areas under the a–t and v–t graphs) to get Δv and Δs.
x = u cos θ · t, y = u sin θ · t − ½gt², vy = u sin θ − gt | v(t) = u + ∫0t a(τ) dτ速度是矢量：题目问"velocity"时要同时给出大小和方向（例如"与水平方向成 11° 向上"）。Velocity is a vector: when asked for "the velocity", give both size and direction (e.g. "at 11° above the horizontal").
| 子题型Sub-type | 识别特征How to spot it | 核心步骤Core step | 真题Past paper |
|---|---|---|---|
| 分段落体Multi-stage fall | "dropped … then sinks / moves at …" | 每段单独算，段间速度接力Each stage separately; velocity links the stages | 2024 Q1(a) |
| 竖直上抛相遇Vertical throws meeting | 两球先后竖直上抛Two balls thrown up at different times | s₁(t) = s₂(t)，第二球用 (t − Δt)s₁(t) = s₂(t), second ball uses (t − Δt) | 2021 Q1(c) |
| 追及 / 超车Chase / overtaking | "passes … overtakes … further down the road" | 同时同地 ⇒ 平均速度相等；v–t 面积相等Same start and finish ⇒ equal average velocity; equal v–t areas | 2016 Q1(f) |
| 斜抛Oblique projectile | "at an angle θ above the horizontal" | x 匀速、y 匀加速，最后合成x uniform, y uniformly accelerated, then combine | 2019 Q1(a) |
| 变加速度Non-uniform acceleration | "acceleration decreases uniformly…""acceleration decreases uniformly…" | Δv = ∫a dt = a–t 图面积Δv = ∫a dt = area under a–t | 2017 Q1(c) |
同类练习（未单独成页）：2023(e) 最后一秒比例的落体、2023(c) 与 2020(f) 的时间账本题、2017(d) 撞墙反弹的抛体。More practice (no page here): 2023(e) last-second ratio in free fall, 2023(c) and 2020(f) time-bookkeeping problems, 2017(d) projectile rebounding from a wall.