Kinematics运动学
Kinematics is the description of motion without yet asking what causes it. Before forces enter the story (that is Dynamics), you learn the precise language of where an object is, how fast it moves, and how its velocity changes. This guide walks the full path from the scalar/vector split (distance vs displacement, speed vs velocity) through acceleration, the four constant-acceleration "SUVAT" equations, the two motion graphs whose slopes and areas carry all the information, free fall under gravity $g$, and finally projectile motion, where a single 2D problem splits cleanly into two independent 1D problems via vector components. Worked examples use real numbers throughout.运动学（kinematics，运动学）只描述运动本身，暂不追问其成因。在力进入故事之前（那是动力学的内容），你要先掌握精确的语言：物体在哪里、运动多快、速度如何变化。本指南从标量／矢量之分（距离与位移、速率与速度）出发，经加速度（acceleration，加速度）、四个匀加速"SUVAT"方程、坡度与面积承载全部信息的两类运动图像、重力 $g$ 下的自由落体（free fall，自由落体），一路推进到抛体运动（projectile motion，抛体运动）——在那里，一道二维问题通过矢量分量干净地拆成两道独立的一维问题。全部例题均用真实数字演算。
How to use this guide如何使用本指南
Kinematics is the first physics unit in every curriculum we map to, and the four curricula agree almost completely on the 1D scope: distance/displacement, speed/velocity, acceleration, the SUVAT equations, motion graphs, and free fall. They diverge on two-dimensional work. US NGSS assessment is explicitly "limited to one-dimensional motion," so the projectile work in §7 sits above the NGSS-assessed floor. Ontario SPH3U (Overall Expectation B3) and Alberta Physics 20 (outcome 20–A1.5k) both name two-dimensional motion as core; BC Physics 11 lists "projectile motion ... angled launch" as Content. The table tells you which sections are core for you right now; each row cites the curriculum document it was checked against.在我们对照的所有大纲中，运动学都是第一个物理单元；四套大纲在一维范围上几乎完全一致：距离／位移、速率／速度、加速度、SUVAT 方程、运动图像与自由落体。它们的分歧在于二维内容。US NGSS 的评估明确"限于一维运动"，因此 §7 的抛体内容高于 NGSS 被评估的下限。安大略 SPH3U（总体期望 B3）与阿尔伯塔 Physics 20（outcome 20–A1.5k）都把二维运动列为核心；BC Physics 11 把"抛体运动……斜抛"列为内容。下表告诉你当前哪些节属于你的核心；每行都注明所依据的课纲文件。
| If you are in…如果你在… | Focus on these sections重点学习 | Defer / lighter可推迟 / 减负 | Source依据 |
|---|---|---|---|
| 🇺🇸 US NGSS HS Physical Science美国 NGSS 物理科学 | §1 through §6 (1D motion, graphs, free fall) — the assessed core under HS-PS2-1§1 至 §6（一维运动、图像、自由落体）—— HS-PS2-1 下被评估的核心 | §7 (2D projectiles): valuable but above the NGSS Assessment Boundary, which is "limited to one-dimensional motion"§7（二维抛体）：很有价值，但高于 NGSS"限于一维运动"的评估边界 | ngss_hs_ps_extract.md — HS-PS2-1 PE + its 1D Assessment Boundary— HS-PS2-1 表现期望及其一维评估边界 |
| 🇨🇦 ON Grade 11 — SPH3U安大略 11 年级 — SPH3U | §1 through §7 in full. SPH3U B3 names "one and two dimensions" explicitly, so projectiles are core§1 至 §7 完整学习。SPH3U B3 明确点名"一维与二维"，故抛体为核心 | Nothing — the full unit is on the Grade 11 syllabus无 — 全单元都在 11 年级大纲内 | science_11-12_physics_extract.md — SPH3U Strand B Overall Expectations B1–B3, Specific Expectation B2.3— SPH3U B 单元总体期望 B1–B3，具体期望 B2.3 |
| 🇨🇦 BC Grade 11 — Physics 11BC 11 年级 — Physics 11 | §1 through §7. Physics 11 Content lists "projectile motion" with "angled launch" elaboration, so §7 is core§1 至 §7。Physics 11 内容含"抛体运动"及"斜抛"细化，故 §7 为核心 | Nothing — BC pairs vectors with right-angle trig from the start, so lean on §7无 — BC 一开始就把矢量与直角三角学配对，故加重 §7 | physics_11-12_extract.md — Physics 11 Content: vectors, uniform/accelerated motion, projectile motion, graphical methods— Physics 11 内容：矢量、匀速/加速运动、抛体运动、图像方法 |
| 🇨🇦 AB Grade 11 — Physics 20阿尔伯塔 11 年级 — Physics 20 | §1 through §7 in full. Physics 20 Unit A GO1 covers all of displacement, velocity, acceleration, and 2D motion via components (20–A1.5k)§1 至 §7 完整学习。Physics 20 A 单元 GO1 覆盖位移、速度、加速度及用分量处理的二维运动（20–A1.5k） | Nothing — AB expects quantitative 2D work; Diploma-style problems reward exact component setup in §7无 — AB 要求定量二维运算；文凭考风格题在 §7 奖励精确的分量设立 | physics_20-30_extract.md — Physics 20 Unit A GO1, knowledge outcomes 20–A1.1k through 20–A1.5k— Physics 20 A 单元 GO1，知识 outcome 20–A1.1k 至 20–A1.5k |
| 🇺🇸 AP / IB feeder trackAP / IB 衔接轨道 | All seven sections plus every going-deeper derivation. AP Physics 1 / C and IB Physics HL all assume fluent SUVAT and projectile decomposition from day one全部 7 节，并完成每个"深入"推导。AP Physics 1 / C 与 IB Physics HL 第一天就默认你熟练 SUVAT 与抛体分解 | Nothing — this unit is the geometric and algebraic foundation that the Dynamics, Energy, and Momentum guides build on无 — 本单元是动力学、能量、动量各指南所依赖的几何与代数基础 | ngss_hs_ps_extract.md — the AP-feeder reads beyond the NGSS 1D floor; see the AP Physics Unit 1 link in "What This Feeds Into"— AP 衔接读到 NGSS 一维下限之上；见"本单元的去向"中的 AP Physics Unit 1 链接 |
Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。
Memorise four things: the definitions $\bar v = \frac{\Delta x}{\Delta t}$ and $\bar a = \frac{\Delta v}{\Delta t}$; the four SUVAT equations (especially $v = v_0 + at$ and $\Delta x = v_0 t + \tfrac{1}{2}at^2$); what slope and area mean on each motion graph (slope of $x$–$t$ is velocity, area under $v$–$t$ is displacement); and that free fall is just SUVAT with $a = g = 9.8\ \mathrm{m/s^2}$ downward. Read every cram-cheat box. Skip the going-deeper derivations.背熟四件事：定义 $\bar v = \frac{\Delta x}{\Delta t}$ 与 $\bar a = \frac{\Delta v}{\Delta t}$；四个 SUVAT 方程（尤其 $v = v_0 + at$ 与 $\Delta x = v_0 t + \tfrac{1}{2}at^2$）；每类运动图像上坡度与面积的含义（$x$–$t$ 图坡度是速度，$v$–$t$ 图下面积是位移）；以及自由落体不过是取 $a = g = 9.8\ \mathrm{m/s^2}$ 向下的 SUVAT。读每个速记框，跳过深入推导。
Always define a positive direction and a coordinate origin before writing any equation; a sign error on $a$ or $v_0$ flips the whole answer. Distinguish average from instantaneous quantities precisely. For projectiles, treat the horizontal and vertical motions as two independent 1D problems linked only by a shared time $t$. Practise reading slope and area off a graph until you can do it without numbers. AB Physics 20 and SPH3U both expect you to derive the SUVAT equations from a velocity–time graph, not just quote them.写任何方程前先定下正方向与坐标原点；$a$ 或 $v_0$ 的符号错误会让整道答案反向。精确区分平均量与瞬时量。处理抛体时，把水平与竖直运动当作两道独立的一维问题，只用共享的时间 $t$ 相连。反复练习从图上读坡度与面积，做到无数字也能读。AB Physics 20 与 SPH3U 都要求你能从速度–时间图推导 SUVAT 方程，而非只是背诵。
Position, Distance and Displacement: Scalars vs Vectors位置、距离与位移：标量与矢量
- Position $x$位置 $x$ — where an object is, measured from a chosen origin along a chosen positive direction.— 物体所在之处，从选定原点沿选定正方向测量。
- Distance距离 — total path length travelled. A scalar (magnitude only, always $\ge 0$).— 走过的总路程。是标量（只有大小，恒 $\ge 0$）。
- Displacement $\Delta x = x_f - x_i$位移 $\Delta x = x_f - x_i$ — the straight-line change in position. A vector (magnitude and direction; can be negative).— 位置的直线变化量。是矢量（有大小和方向；可为负）。
20–A1.2k: "compare and contrast scalar and vector quantities"); BC Physics 11 lists "vector and scalar quantities" as its first motion Content bullet.最易踩坑的一点：若你向东走 $3$ m 再向西走回 $3$ m，距离是 $6$ m，但位移为 $0$。阿尔伯塔直接陈述此点（20–A1.2k："比较标量与矢量"）；BC Physics 11 把"矢量与标量"列为运动内容的第一条。
A runner jogs $400$ m east, then turns and jogs $150$ m west along the same road. Taking east as positive, find (a) the total distance and (b) the displacement.一名跑者向东慢跑 $400$ m，随后掉头沿同一条路向西慢跑 $150$ m。取向东为正，求 (a) 总距离与 (b) 位移。
Distance — add path lengths.距离 — 路程相加。 Distance is a scalar, so directions do not cancel: $400 + 150 = 550$ m.距离是标量，方向不抵消：$400 + 150 = 550$ m。
Displacement — net change in position.位移 — 位置的净变化。
$$ \Delta x = (+400) + (-150) = +250 \text{ m (east)}. $$State the answer with direction.作答时带方向。 Distance $= 550$ m; displacement $= 250$ m east. The magnitudes differ because the runner doubled back — whenever the path is not a straight one-way line, distance $>$ |displacement|.距离 $= 550$ m；位移 $= 250$ m 向东。两者大小不同，因为跑者折返了 — 只要路径不是直线单向，就有 距离 $>$ |位移|。
Going deeper — why physics insists on a signed coordinate, not just "forward/back"深入 — 物理为何坚持用带符号的坐标，而非仅"前／后"
Everyday language treats "distance moved" as enough. Physics replaces it with a coordinate $x$ measured along an axis with a chosen origin and a chosen positive direction, because every later quantity is built as a difference or rate of change of $x$. Velocity is $\frac{\Delta x}{\Delta t}$; acceleration is $\frac{\Delta v}{\Delta t}$. If $x$ were unsigned, a reversal of motion could not be expressed, and the SUVAT equations of §4 would lose the sign bookkeeping that makes them work.日常语言把"移动的距离"当作足够。物理用沿轴测量、带选定原点与正方向的坐标 $x$ 取而代之，因为后续每个量都是 $x$ 的差或变化率。速度是 $\frac{\Delta x}{\Delta t}$；加速度是 $\frac{\Delta v}{\Delta t}$。若 $x$ 无符号，运动的反向便无法表达，§4 的 SUVAT 方程也会失去使其成立的符号记账。
Concretely: choose east $= +$. Then a velocity of $-2\ \mathrm{m/s}$ means "moving west at $2\ \mathrm{m/s}$," and an acceleration of $-3\ \mathrm{m/s^2}$ while moving west (negative velocity) means speeding up, not slowing down. The sign of $a$ alone never tells you "faster or slower" — you must compare the signs of $a$ and $v$. This is the single most common conceptual error in the unit, and every later section depends on getting it right here.具体地说：取向东 $= +$。则速度 $-2\ \mathrm{m/s}$ 表示"以 $2\ \mathrm{m/s}$ 向西运动"，而向西运动（速度为负）时加速度 $-3\ \mathrm{m/s^2}$ 表示加速，而非减速。仅凭 $a$ 的符号永远说不清"变快还是变慢"——你必须比较 $a$ 与 $v$ 的符号。这是本单元最常见的概念错误，后续每节都依赖在此处理对它。
Velocity and Speed: Average vs Instantaneous速度与速率：平均与瞬时
- Velocity vs speed.速度与速率。 Velocity is a vector (uses displacement); speed is a scalar (uses distance). On a round trip, average velocity can be $0$ while average speed is not.速度是矢量（用位移）；速率是标量（用距离）。往返行程中，平均速度可为 $0$，平均速率却不为零。
- Average vs instantaneous.平均与瞬时。 Average uses a whole interval $\Delta t$; instantaneous velocity is the limit as $\Delta t \to 0$ — the slope of the position–time graph at one instant (the speedometer reading).平均量用整段区间 $\Delta t$；瞬时速度是 $\Delta t \to 0$ 的极限 — 即位置–时间图在某一瞬间的坡度（速度表读数）。
- Units.单位。 SI is $\mathrm{m/s}$. To convert $\mathrm{km/h} \to \mathrm{m/s}$, divide by $3.6$.国际单位是 $\mathrm{m/s}$。$\mathrm{km/h} \to \mathrm{m/s}$ 除以 $3.6$。
20–A1.1k requires you to "define, qualitatively and quantitatively, displacement, velocity and acceleration"; the average/instantaneous distinction is the gateway to the derivative used at AP/IB level.阿尔伯塔 20–A1.1k 要求你"定性与定量地定义位移、速度与加速度"；平均／瞬时之分是通往 AP／IB 级所用导数的门槛。
A car drives $60$ km east in $1.0$ h, then $40$ km west in $0.5$ h. Taking east as positive, find the average velocity and the average speed for the whole trip.一辆车用 $1.0$ h 向东行 $60$ km，再用 $0.5$ h 向西行 $40$ km。取向东为正，求整段行程的平均速度与平均速率。
Net displacement and total time.净位移与总时间。 $\Delta x = (+60) + (-40) = +20$ km. Total time $\Delta t = 1.0 + 0.5 = 1.5$ h.$\Delta x = (+60) + (-40) = +20$ km。总时间 $\Delta t = 1.0 + 0.5 = 1.5$ h。
Average velocity (vector).平均速度（矢量）。
$$ \bar v = \frac{\Delta x}{\Delta t} = \frac{+20 \text{ km}}{1.5 \text{ h}} \approx +13.3 \text{ km/h (east)}. $$Average speed (scalar).平均速率（标量）。
$$ \text{avg speed} = \frac{60 + 40}{1.5} = \frac{100}{1.5} \approx 66.7 \text{ km/h}. $$Why they differ.两者为何不同。 Average speed uses total distance $100$ km; average velocity uses the much smaller net displacement $20$ km. They agree only when motion never reverses.平均速率用总距离 $100$ km；平均速度用小得多的净位移 $20$ km。只有运动从不反向时两者才相等。
Going deeper — instantaneous velocity as the limit of average velocity深入 — 瞬时速度作为平均速度的极限
Average velocity over an interval is $\bar v = \frac{\Delta x}{\Delta t}$. Shrink the interval around a single instant $t$: as $\Delta t \to 0$, the ratio approaches a definite value, the instantaneous velocity某区间上的平均速度为 $\bar v = \frac{\Delta x}{\Delta t}$。把区间在某一瞬间 $t$ 周围收缩：当 $\Delta t \to 0$ 时，该比值趋于一个确定值，即瞬时速度
$$ v(t) = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}. $$Geometrically, this is the slope of the tangent line to the position–time graph at $t$ (see §5). At HS level you read this slope graphically; at AP Physics C and IB Physics HL it becomes the derivative $\frac{dx}{dt}$ outright. The same limit turns average acceleration into instantaneous acceleration in §3. Recognising velocity as "slope of $x$ vs $t$" and acceleration as "slope of $v$ vs $t$" is the conceptual spine of the whole unit.几何上，这是位置–时间图在 $t$ 处切线的坡度（见 §5）。在高中阶段你用图像读出此坡度；在 AP Physics C 与 IB Physics HL 中它直接成为导数 $\frac{dx}{dt}$。同一极限在 §3 把平均加速度变为瞬时加速度。把速度认作"$x$ 对 $t$ 的坡度"、加速度认作"$v$ 对 $t$ 的坡度"，是整个单元的概念主干。
Acceleration: the Rate of Change of Velocity加速度：速度的变化率
- Sign rule.符号法则。 If $a$ and $v$ have the same sign, the object speeds up. If they have opposite signs, it slows down. "Negative acceleration" does not mean "slowing down."若 $a$ 与 $v$ 符号相同，物体加速。若符号相反，则减速。"负加速度"不等于"减速"。
- Constant (uniform) acceleration恒定（匀）加速度 — the case that the SUVAT equations of §4 assume. Free fall is the headline example ($a = g$).— §4 的 SUVAT 方程所假设的情形。自由落体是头号例子（$a = g$）。
- Deceleration减速 is just acceleration opposite to the velocity; it is not a separate quantity.不过是与速度反向的加速度；并非另一种量。
20–A1.3k asks you to "explain ... uniform and uniformly accelerated motion." The uniform-acceleration assumption is what makes the next section's four equations valid.阿尔伯塔 20–A1.3k 要求你"解释……匀速与匀加速运动"。正是匀加速假设使下一节四个方程成立。
A car moving east at $25\ \mathrm{m/s}$ brakes to $10\ \mathrm{m/s}$ over $3.0$ s, still moving east. Taking east as positive, find the acceleration and state whether the car is speeding up or slowing down.一辆向东以 $25\ \mathrm{m/s}$ 行驶的车在 $3.0$ s 内刹车到 $10\ \mathrm{m/s}$，仍向东。取向东为正，求加速度并说明车在加速还是减速。
Apply the definition.套用定义。
$$ \bar a = \frac{v_f - v_i}{\Delta t} = \frac{10 - 25}{3.0} = \frac{-15}{3.0} = -5.0\ \mathrm{m/s^2}. $$Interpret the sign.解读符号。 Velocity is positive (east) but acceleration is negative (west). Opposite signs $\Rightarrow$ the car is slowing down. The acceleration points west, opposing the motion — this is the braking force at work.速度为正（东），加速度为负（西）。符号相反 $\Rightarrow$ 车在减速。加速度指向西，与运动相反 — 这正是刹车力的作用。
Sanity-check.合理性核验。 A magnitude of $5\ \mathrm{m/s^2}$ means the speed drops by $5\ \mathrm{m/s}$ each second: $25 \to 20 \to 15 \to 10$ over three seconds. ✓大小 $5\ \mathrm{m/s^2}$ 表示速率每秒降 $5\ \mathrm{m/s}$：三秒内 $25 \to 20 \to 15 \to 10$。✓
The Kinematic Equations (Constant Acceleration / SUVAT)运动学方程（匀加速 / SUVAT）
Symbols: $v_0$ initial velocity, $v$ final velocity, $a$ acceleration, $t$ time, $\Delta x$ displacement. Each equation omits exactly one of the five.符号：$v_0$ 初速度，$v$ 末速度，$a$ 加速度，$t$ 时间，$\Delta x$ 位移。每个方程恰好省略五者之一。
$$ v = v_0 + a t \qquad\qquad \Delta x = v_0 t + \tfrac{1}{2} a t^2 $$ $$ v^2 = v_0^2 + 2 a\,\Delta x \qquad\qquad \Delta x = \tfrac{1}{2}(v_0 + v)\, t $$- How to choose.如何选式。 List the three knowns and the unknown; pick the equation that contains those four and omits the fifth. Equation 3 ($v^2 = v_0^2 + 2a\Delta x$) is the time-free one.列出三个已知与一个未知；选恰含这四者、省去第五者的方程。式 3（$v^2 = v_0^2 + 2a\Delta x$）是不含时间的那个。
- Sign discipline.符号纪律。 Fix one positive direction. Every $v_0$, $v$, $a$, $\Delta x$ takes a sign from that choice. This is where most marks are lost.定下一个正方向。每个 $v_0$、$v$、$a$、$\Delta x$ 都依此取符号。失分多在此处。
B2.3 requires you to "derive ... the equations for displacement" from a velocity–time graph — see the derivation in the going-deeper box.安大略 B2.3 要求你从速度–时间图"推导……位移方程"——见深入框中的推导。
A car travelling at $30\ \mathrm{m/s}$ brakes with a constant acceleration of $-6.0\ \mathrm{m/s^2}$. How far does it travel before stopping?一辆以 $30\ \mathrm{m/s}$ 行驶的车以恒定加速度 $-6.0\ \mathrm{m/s^2}$ 刹车。停下前行驶多远？
List knowns and unknown.列已知与未知。 $v_0 = 30$, $v = 0$ (stops), $a = -6.0$, find $\Delta x$. Time is not given and not asked — use the time-free equation.$v_0 = 30$，$v = 0$（停下），$a = -6.0$，求 $\Delta x$。时间未知也未问 — 用不含时间的方程。
$$ v^2 = v_0^2 + 2 a\,\Delta x \;\Longrightarrow\; 0 = 30^2 + 2(-6.0)\,\Delta x. $$Solve.求解。
$$ \Delta x = \frac{-30^2}{2(-6.0)} = \frac{-900}{-12} = 75 \text{ m}. $$Answer with units.带单位作答。 The car travels $75$ m before stopping. (Sanity-check: stopping distance scales with $v_0^2$ — doubling the speed quadruples the distance, the reason speed limits matter so much.)车在停下前行驶 $75$ m。（合理性核验：刹车距离与 $v_0^2$ 成正比 — 速度加倍则距离变四倍，这正是限速如此重要的原因。）
Going deeper — deriving the SUVAT equations from a velocity–time graph (ON B2.3, AB 20–A1.3k)深入 — 从速度–时间图推导 SUVAT 方程（ON B2.3，AB 20–A1.3k）
For constant acceleration, the velocity–time graph is a straight line from $v_0$ to $v$ over time $t$. Two facts about that line generate all four equations.匀加速时，速度–时间图是一条从 $v_0$ 到 $v$、历时 $t$ 的直线。关于这条直线的两个事实即可生成全部四个方程。
1. Slope $=$ acceleration.1. 坡度 $=$ 加速度。 The slope of the line is $\frac{v - v_0}{t} = a$, which rearranges to $v = v_0 + at$. (Equation 1.)直线坡度为 $\frac{v - v_0}{t} = a$，整理得 $v = v_0 + at$。（式 1。）
2. Area $=$ displacement.2. 面积 $=$ 位移。 The area under the line is a trapezoid: $\Delta x = \tfrac{1}{2}(v_0 + v)t$. (Equation 4.) Substituting $v = v_0 + at$ into it gives $\Delta x = \tfrac{1}{2}(v_0 + v_0 + at)t = v_0 t + \tfrac{1}{2}at^2$. (Equation 2.)直线下面积是一个梯形：$\Delta x = \tfrac{1}{2}(v_0 + v)t$。（式 4。）代入 $v = v_0 + at$ 得 $\Delta x = \tfrac{1}{2}(v_0 + v_0 + at)t = v_0 t + \tfrac{1}{2}at^2$。（式 2。）
3. Eliminate $t$.3. 消去 $t$。 From Equation 1, $t = (v - v_0)/a$. Substituting into Equation 4 and simplifying yields $v^2 = v_0^2 + 2a\Delta x$. (Equation 3, the time-free one.)由式 1，$t = (v - v_0)/a$。代入式 4 并化简得 $v^2 = v_0^2 + 2a\Delta x$。（式 3，即不含时间者。）
This graph-area / graph-slope reasoning is exactly what Ontario B2.3 names, and it is why §5 (motion graphs) is not a side topic but the engine behind these formulas.这种"图像面积／图像坡度"的推理正是安大略 B2.3 所点名的，也是为何 §5（运动图像）不是旁支，而是这些公式背后的引擎。
Motion Graphs: Slopes and Areas运动图像：坡度与面积
- Position–time ($x$ vs $t$):位置–时间（$x$ 对 $t$）： the slope is the velocity. Steeper $=$ faster; a horizontal line $=$ at rest; a downward slope $=$ moving in the negative direction.坡度就是速度。越陡越快；水平线 $=$ 静止；下行坡度 $=$ 向负方向运动。
- Velocity–time ($v$ vs $t$):速度–时间（$v$ 对 $t$）： the slope is the acceleration, and the area between the line and the time axis is the displacement.坡度是加速度，曲线与时间轴之间的面积是位移。
- Sign of area.面积的符号。 Area below the time axis counts as negative displacement.时间轴下方的面积计为负位移。
An object starts at rest and its velocity rises linearly to $12\ \mathrm{m/s}$ over the first $4.0$ s, then stays constant at $12\ \mathrm{m/s}$ for the next $6.0$ s. Find (a) the acceleration during the first phase and (b) the total displacement over the $10.0$ s.某物体从静止出发，速度在前 $4.0$ s 内线性升到 $12\ \mathrm{m/s}$，随后 $6.0$ s 保持 $12\ \mathrm{m/s}$ 不变。求 (a) 第一阶段的加速度与 (b) 这 $10.0$ s 的总位移。
(a) Acceleration $=$ slope of the rising segment.(a) 加速度 $=$ 上升段的坡度。
$$ a = \frac{\Delta v}{\Delta t} = \frac{12 - 0}{4.0} = 3.0\ \mathrm{m/s^2}. $$(b) Displacement $=$ area under the whole graph.(b) 位移 $=$ 整张图下的面积。 Phase 1 is a triangle; phase 2 is a rectangle.第一阶段是三角形；第二阶段是矩形。
$$ \Delta x = \underbrace{\tfrac{1}{2}(4.0)(12)}_{\text{triangle}} + \underbrace{(6.0)(12)}_{\text{rectangle}} = 24 + 72 = 96 \text{ m}. $$Answer.作答。 Acceleration $3.0\ \mathrm{m/s^2}$ in phase 1; total displacement $96$ m. Notice no SUVAT formula was needed — the graph carried all the information directly.第一阶段加速度 $3.0\ \mathrm{m/s^2}$；总位移 $96$ m。注意全程未用 SUVAT 公式 — 图像直接承载了全部信息。
Free Fall and Vertical Motion under Gravity自由落体与重力下的竖直运动
- $g \approx 9.8\ \mathrm{m/s^2}$$g \approx 9.8\ \mathrm{m/s^2}$ (often $9.81$, sometimes rounded to $10$), directed downward, the same for all masses when air resistance is neglected.（常取 $9.81$，有时约为 $10$），方向竖直向下，忽略空气阻力时对所有质量相同。
- Mass-independence.与质量无关。 A feather and a hammer fall identically in vacuum — $g$ does not depend on mass.真空中羽毛与铁锤下落完全相同 — $g$ 不依赖质量。
- At the top of a thrown object's flight,在抛起物体飞行的最高点， $v = 0$ for an instant, but $a = g$ still points down. Velocity zero does not mean acceleration zero.瞬间 $v = 0$，但 $a = g$ 仍指向下。速度为零不代表加速度为零。
A ball is thrown straight up at $19.6\ \mathrm{m/s}$. Using $g = 9.8\ \mathrm{m/s^2}$ and up as positive, find (a) the time to reach the highest point and (b) the maximum height above the launch point. Ignore air resistance.一只球以 $19.6\ \mathrm{m/s}$ 竖直上抛。取 $g = 9.8\ \mathrm{m/s^2}$ 且向上为正，求 (a) 到达最高点的时间与 (b) 高于抛点的最大高度。忽略空气阻力。
(a) At the top, $v = 0$.(a) 在最高点 $v = 0$。 Use $v = v_0 - gt$ with $v = 0$:用 $v = v_0 - gt$ 并取 $v = 0$：
$$ 0 = 19.6 - 9.8\,t \;\Longrightarrow\; t = \frac{19.6}{9.8} = 2.0 \text{ s}. $$(b) Maximum height.(b) 最大高度。 Use the time-free equation $v^2 = v_0^2 - 2g\,\Delta y$ with $v = 0$:用不含时间的 $v^2 = v_0^2 - 2g\,\Delta y$ 并取 $v = 0$：
$$ 0 = 19.6^2 - 2(9.8)\,\Delta y \;\Longrightarrow\; \Delta y = \frac{19.6^2}{2(9.8)} = \frac{384.16}{19.6} = 19.6 \text{ m}. $$Sanity-check.合理性核验。 By symmetry the ball takes another $2.0$ s to fall back, returning at $-19.6\ \mathrm{m/s}$ — same speed, opposite direction. At the very top, $v = 0$ but $a = g$ still acts downward, which is exactly why the ball does not hang there. ✓由对称性，球再用 $2.0$ s 落回，返回时速度为 $-19.6\ \mathrm{m/s}$ — 速率相同、方向相反。在最高点 $v = 0$ 但 $a = g$ 仍向下作用，这正是球不会停留在那里的原因。✓
Projectile Motion and 2D Kinematics Honors — US NGSS (1D-assessed)抛体运动与二维运动学 荣誉 — US NGSS（仅评估一维）
Split a launch velocity $v_0$ at angle $\theta$ into components, then treat each axis separately.把以角 $\theta$ 发射的初速度 $v_0$ 分解为分量，再对每个轴分别处理。
$$ v_{0x} = v_0 \cos\theta, \qquad v_{0y} = v_0 \sin\theta. $$- Horizontal ($x$):水平（$x$）： no horizontal force, so $a_x = 0$ and $v_x = v_{0x}$ stays constant: $\Delta x = v_{0x}\,t$.无水平力，故 $a_x = 0$，$v_x = v_{0x}$ 保持不变：$\Delta x = v_{0x}\,t$。
- Vertical ($y$):竖直（$y$）： free fall, $a_y = -g$: $\Delta y = v_{0y}\,t - \tfrac{1}{2}g t^2$.自由落体，$a_y = -g$：$\Delta y = v_{0y}\,t - \tfrac{1}{2}g t^2$。
- The link.联系。 Time $t$ is the only quantity shared by both axes. Solve one axis for $t$, then feed it into the other.时间 $t$ 是两轴唯一共享的量。在一个轴上解出 $t$，再代入另一轴。
A ball rolls off a table $1.25$ m high with a horizontal speed of $3.0\ \mathrm{m/s}$. Using $g = 9.8\ \mathrm{m/s^2}$, find (a) the time to hit the floor and (b) the horizontal distance travelled. Ignore air resistance.一只球以 $3.0\ \mathrm{m/s}$ 的水平速率从 $1.25$ m 高的桌面滚下。取 $g = 9.8\ \mathrm{m/s^2}$，求 (a) 落地时间与 (b) 水平方向行进的距离。忽略空气阻力。
(a) Time comes from the vertical axis alone.(a) 时间仅由竖直轴决定。 The initial vertical velocity is zero ($v_{0y} = 0$, a horizontal launch). With down as positive for this part:初始竖直速度为零（$v_{0y} = 0$，水平抛出）。本小问取向下为正：
$$ \Delta y = \tfrac{1}{2} g t^2 \;\Longrightarrow\; 1.25 = \tfrac{1}{2}(9.8)t^2 \;\Longrightarrow\; t^2 = \frac{2(1.25)}{9.8} \approx 0.255 \;\Longrightarrow\; t \approx 0.505 \text{ s}. $$(b) Horizontal distance uses that shared time.(b) 水平距离用这一共享时间。 Horizontal velocity is constant ($a_x = 0$):水平速度恒定（$a_x = 0$）：
$$ \Delta x = v_{0x}\,t = 3.0 \times 0.505 \approx 1.52 \text{ m}. $$The key idea.关键思想。 The horizontal speed had no effect on the fall time — a ball dropped and a ball launched horizontally from the same height hit the floor together. The two axes are independent, linked only through $t$.水平速率对落地时间毫无影响 — 自同一高度静止释放的球与水平抛出的球同时落地。两轴相互独立，只通过 $t$ 相连。
Going deeper — range of an angled projectile on level ground (AB 20–A1.5k)深入 — 平地上斜抛的射程（AB 20–A1.5k）
Launch from ground level at speed $v_0$ and angle $\theta$. Take up as positive, $a_y = -g$, $a_x = 0$.从地面以速率 $v_0$、角 $\theta$ 发射。取向上为正，$a_y = -g$，$a_x = 0$。
Time of flight.飞行时间。 The projectile returns to $\Delta y = 0$. From $\Delta y = v_{0y}t - \tfrac{1}{2}gt^2 = 0$, factoring gives $t(v_{0y} - \tfrac{1}{2}gt) = 0$, so the non-trivial root is抛体回到 $\Delta y = 0$。由 $\Delta y = v_{0y}t - \tfrac{1}{2}gt^2 = 0$，提取因式得 $t(v_{0y} - \tfrac{1}{2}gt) = 0$，故非平凡根为
$$ t_{\text{flight}} = \frac{2 v_{0y}}{g} = \frac{2 v_0 \sin\theta}{g}. $$Range.射程。 Horizontal distance is $R = v_{0x}\,t_{\text{flight}} = v_0\cos\theta \cdot \frac{2 v_0 \sin\theta}{g}$. Using $2\sin\theta\cos\theta = \sin 2\theta$,水平距离为 $R = v_{0x}\,t_{\text{flight}} = v_0\cos\theta \cdot \frac{2 v_0 \sin\theta}{g}$。利用 $2\sin\theta\cos\theta = \sin 2\theta$，
$$ R = \frac{v_0^2 \sin 2\theta}{g}. $$This compact result shows the maximum range occurs at $\theta = 45^{\circ}$ (where $\sin 2\theta = 1$), and that complementary angles such as $30^{\circ}$ and $60^{\circ}$ give the same range. The whole derivation is nothing but §4's SUVAT applied separately on each axis — exactly the "vector components" method Alberta names in outcome 20–A1.5k, and the foundation AP Physics and IB Physics HL build their projectile and 2D-motion problems on.这个紧凑结果表明最大射程出现在 $\theta = 45^{\circ}$（此时 $\sin 2\theta = 1$），且互余角如 $30^{\circ}$ 与 $60^{\circ}$ 给出相同射程。整个推导不过是把 §4 的 SUVAT 在每个轴上分别应用 — 正是阿尔伯塔在 outcome 20–A1.5k 中所称的"矢量分量"法，也是 AP Physics 与 IB Physics HL 构建抛体与二维运动问题的基础。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Define positive first.先定正方向。 Write down which direction is positive and where the origin is before any equation. Then assign signs to $v_0$, $v$, $a$, $\Delta x$ consistently. The single most common error in this unit is a sign flip.写任何方程前，先写明哪个方向为正、原点在哪。再一致地给 $v_0$、$v$、$a$、$\Delta x$ 赋符号。本单元最常见的错误就是符号反向。
- List knowns, then pick the equation.列已知，再选式。 Write the three knowns and the unknown; choose the SUVAT equation containing exactly those four. If time is neither given nor asked, reach for $v^2 = v_0^2 + 2a\Delta x$.写下三个已知与一个未知；选恰含这四者的 SUVAT 方程。若时间既不给也不问，就用 $v^2 = v_0^2 + 2a\Delta x$。
- Carry units, answer with direction.带单位，作答带方向。 Convert $\mathrm{km/h}$ to $\mathrm{m/s}$ early (divide by $3.6$). For vector answers, state the direction ("$250$ m east"), not just the magnitude.尽早把 $\mathrm{km/h}$ 换成 $\mathrm{m/s}$（除以 $3.6$）。矢量答案要说明方向（"$250$ m 向东"），不只是大小。
- Sign of $a$ never means "slowing down."$a$ 的符号绝不等于"减速"。 Compare the signs of $a$ and $v$: same $=$ speeding up, opposite $=$ slowing down. Negative acceleration on a negative velocity speeds the object up.比较 $a$ 与 $v$ 的符号：同号加速，异号减速。负速度上的负加速度使物体加速。
- Slope and area are your shortcut.坡度与面积是你的捷径。 Slope of $x$–$t$ is $v$; slope of $v$–$t$ is $a$; area under $v$–$t$ is $\Delta x$. Many "hard" graph questions need no formula at all.$x$–$t$ 坡度是 $v$；$v$–$t$ 坡度是 $a$；$v$–$t$ 下面积是 $\Delta x$。许多"难"图像题根本不需公式。
- At the top, $v = 0$ but $a = g$.在最高点 $v = 0$ 但 $a = g$。 A thrown object stops for an instant at its peak, yet gravity keeps acting. Velocity zero never implies acceleration zero.抛起物在顶点瞬间停止，但重力持续作用。速度为零绝不意味加速度为零。
- Split into two 1D problems.拆成两道一维问题。 Resolve $v_0$ into $v_{0x} = v_0\cos\theta$ and $v_{0y} = v_0\sin\theta$, then run horizontal ($a_x = 0$) and vertical ($a_y = -g$) motions separately.把 $v_0$ 分解为 $v_{0x} = v_0\cos\theta$ 与 $v_{0y} = v_0\sin\theta$，再分别处理水平（$a_x = 0$）与竖直（$a_y = -g$）运动。
- Time is the bridge.时间是桥梁。 The two axes share only $t$. Usually the vertical axis sets the flight time; feed that $t$ into the horizontal axis for range.两轴只共享 $t$。通常由竖直轴确定飞行时间；把该 $t$ 代入水平轴求射程。
- Horizontal speed never changes the fall time.水平速率从不改变落地时间。 A dropped object and a horizontally launched one from the same height land together — the independence of the axes is a favourite exam point.自同一高度静止释放与水平抛出的物体同时落地 — 两轴的独立性是常考点。
- Round at the very end.最后一步再四舍五入。 Carry extra digits through intermediate steps; round only the final number to the precision the question asks.中间步骤多留几位；仅在最终答案处按题目要求精度四舍五入。
- Sanity-check magnitudes.用数量级核验。 Stopping distance grows with $v_0^2$; free-fall speed grows as $gt$; a $45^{\circ}$ launch maximises range. If a number looks wrong, recheck the sign convention first.刹车距离随 $v_0^2$ 增长；自由落体速度随 $gt$ 增长；$45^{\circ}$ 发射射程最大。若数字可疑，先复查符号约定。
- State $g$ explicitly.明确写出 $g$。 Write whether you used $9.8$, $9.81$, or $10\ \mathrm{m/s^2}$; the marker matches your rounding to it.写明你用了 $9.8$、$9.81$ 还是 $10\ \mathrm{m/s^2}$；评分者据此核对你的舍入。
Flashcards闪卡
Practice Quiz综合测验
Readiness Checklist准备就绪清单
Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。
- Classify each of distance, displacement, speed, velocity, and acceleration as a scalar or a vector, and explain why displacement can be smaller than distance. 🇨🇦 AB 20–A1.2k把距离、位移、速率、速度、加速度逐一归为标量或矢量，并解释位移为何可小于距离。🇨🇦 AB 20–A1.2k
- Compute average velocity and average speed for a multi-leg trip, and explain when the two differ.对多段行程计算平均速度与平均速率，并解释两者何时不同。
- State the average-acceleration definition and use the sign rule (compare signs of $a$ and $v$) to decide whether an object is speeding up or slowing down.陈述平均加速度定义，并用符号法则（比较 $a$ 与 $v$ 的符号）判断物体在加速还是减速。
- Write the four SUVAT equations from memory and choose the right one given any three of $\{v_0, v, a, t, \Delta x\}$.默写四个 SUVAT 方程，并在已知 $\{v_0, v, a, t, \Delta x\}$ 中任意三个时选对方程。
- Solve a constant-acceleration problem with correct sign conventions, including the time-free equation $v^2 = v_0^2 + 2a\Delta x$.用正确的符号约定求解匀加速问题，包括不含时间的 $v^2 = v_0^2 + 2a\Delta x$。
- Read velocity off a position–time graph (slope) and read acceleration and displacement off a velocity–time graph (slope and area). 🇨🇦 BC Physics 11 graphical methods从位置–时间图读速度（坡度），从速度–时间图读加速度与位移（坡度与面积）。🇨🇦 BC Physics 11 图像方法
- Derive the SUVAT equations from a velocity–time graph (slope $=a$, trapezoid area $=\Delta x$, eliminate $t$). 🇨🇦 ON SPH3U B2.3从速度–时间图推导 SUVAT 方程（坡度 $=a$，梯形面积 $=\Delta x$，消去 $t$）。🇨🇦 ON SPH3U B2.3
- Solve free-fall problems using $a = g = 9.8\ \mathrm{m/s^2}$ with a consistent up-positive convention, including a ball thrown straight up. 🇺🇸 NGSS HS-PS2-1 ("a falling object")用 $a = g = 9.8\ \mathrm{m/s^2}$、一致的向上为正约定求解自由落体问题，包括竖直上抛的球。🇺🇸 NGSS HS-PS2-1（"下落物体"）
- Explain why an object thrown straight up has zero velocity but non-zero acceleration ($g$) at its highest point.解释竖直上抛物体为何在最高点速度为零但加速度非零（为 $g$）。
- Honors (US) Resolve a launch velocity into components ($v_0\cos\theta$, $v_0\sin\theta$) and solve a projectile problem as two independent 1D motions linked by a shared time $t$. 🇨🇦 ON B3 / BC P11 / AB 20–A1.5k coreHonors（US） 把发射速度分解为分量（$v_0\cos\theta$、$v_0\sin\theta$），并把抛体问题当作由共享时间 $t$ 相连的两道独立一维运动求解。🇨🇦 ON B3 / BC P11 / AB 20–A1.5k 核心
- Honors (US) Explain that horizontal speed does not affect fall time, and (going deeper) derive the level-ground range $R = \frac{v_0^2\sin 2\theta}{g}$, noting that $45^{\circ}$ maximises range.Honors（US） 解释水平速率不影响落地时间，并（深入）推导平地射程 $R = \frac{v_0^2\sin 2\theta}{g}$，指出 $45^{\circ}$ 使射程最大。
What This Feeds Into本单元的去向
Kinematics is the descriptive foundation for every later mechanics topic. The next HS Physics unit, Dynamics, asks what causes the acceleration you described here (Newton's laws); Work-Energy and Momentum reuse the same velocity and displacement quantities. The vector-component method of §7 reappears everywhere a force or velocity must be resolved on an incline or in a plane. The cross-references below point at AP and IB units already shipped in this repo, and at the HS Math guide that grounds the trigonometry §7 depends on.运动学是后续每个力学主题的描述性基础。下一个 HS Physics 单元《动力学》追问你在此处所描述的加速度由何引起（牛顿定律）；功-能与动量复用同样的速度与位移量。§7 的矢量分量法在任何需要把力或速度在斜面或平面上分解之处都会再现。下方链接指向本仓库已有的 AP 与 IB 单元，以及为 §7 所依赖的三角学打基础的 HS Math 指南。
Within High School Physics.在 HS Physics 内部。
The Dynamics guide (Forces and Newton's Laws) takes the acceleration $a$ you computed here and ties it to net force via $F = ma$. Work, Energy and Power reuses displacement and velocity to build $W = F\Delta x$ and $KE = \tfrac{1}{2}mv^2$. Momentum and Collisions reuses velocity directly in $p = mv$. Circular Motion treats centripetal acceleration as a special 2D case of the kinematics you met in §7.《动力学》（力与牛顿定律）把你在此算出的加速度 $a$ 通过 $F = ma$ 与净力相连。《功、能与功率》复用位移与速度构建 $W = F\Delta x$ 与 $KE = \tfrac{1}{2}mv^2$。《动量与碰撞》直接在 $p = mv$ 中复用速度。《圆周运动》把向心加速度视为 §7 所遇二维运动学的一个特例。
Across the AP, IB, and HS Math feeders in this repo.本仓库中的 AP、IB 与 HS Math 衔接单元。
If you are aiming for AP Physics 1 or C, the SUVAT fluency and projectile decomposition here are assumed from the first week; AP adds calculus definitions of velocity and acceleration on top. For IB Physics HL, Topic A1 (Kinematics) picks up exactly this scope and extends it with uncertainty analysis and graphing. The component method in §7 is the same one used throughout the AP Physics mechanics units whenever a vector is resolved on an incline.备考 AP Physics 1 或 C：第一周就默认你熟练此处的 SUVAT 与抛体分解；AP 在其上加入速度与加速度的微积分定义。备考 IB Physics HL：主题 A1（运动学）正是接续此范围，并以不确定度分析与作图加以拓展。§7 的分量法与 AP Physics 力学各单元在斜面上分解矢量时所用方法相同。