Unit A.1: Kinematics单元 A.1：运动学
The opening unit of Theme A "Space, time and motion". Distance and displacement, the five suvat equations of constant-acceleration motion, motion graphs, free fall, projectile motion, vector decomposition, and relative velocity. Every later mechanics unit (forces, energy, momentum, rigid bodies, relativity) builds on the kinematic language introduced here. The HL extension covers fluid resistance and terminal velocity, plus a closer look at relative velocity in moving reference frames.主题 A"空间、时间与运动"的开篇。距离与位移、匀加速运动的五个 suvat 方程、运动图像、自由落体、抛体运动、矢量分解，以及相对速度。后续所有力学单元（力、能量、动量、刚体、相对论）都建立在本单元的运动学语言之上。HL 扩展涵盖流体阻力与收尾速度，并深入讲解动参考系中的相对速度。
How to use this guide本指南使用说明
A.1 is the vocabulary unit of mechanics. The equations are short and the algebra is light; the marks come from precise vector reasoning (displacement is not distance, velocity is not speed) and from picking the right suvat equation given the four pieces of information the question hands you. Train the language alongside the algebra.A.1 是力学的词汇单元。公式很短、代数也轻；分数来自精确的矢量推理（位移不等于路程，速度不等于速率）以及在题目给出的四条信息中选对 suvat 方程。术语与代数一起练。
Memorise the five suvat equations and the projectile decomposition $v_x = v_0 \cos\theta$, $v_y = v_0 \sin\theta - g t$. Pick the suvat that contains your four known quantities (one of $s, u, v, a, t$ is always unknown). Velocity is the slope of an $x$-$t$ graph; displacement is the signed area under a $v$-$t$ graph.
背熟五个 suvat 方程与斜抛分解 $v_x = v_0 \cos\theta$、$v_y = v_0 \sin\theta - g t$。在题目给出的四个量中挑选合适的 suvat（$s, u, v, a, t$ 中总有一个未知）。速度是 $x$-$t$ 图的斜率；位移是 $v$-$t$ 图的有符号面积。
Be fluent in deriving suvat from $v = u + at$ and the mean-velocity identity $s = \tfrac{1}{2}(u + v) t$. State the sign convention every time you set up a problem. For projectile and relative-velocity items, draw a labelled diagram with axes before computing. Understand why $v_{\text{terminal}}$ exists (drag balances weight) and what shape a $v$-$t$ graph takes when drag is present.
能从 $v = u + at$ 与平均速度恒等式 $s = \tfrac{1}{2}(u + v) t$ 推出全部 suvat。每次列方程前都要写明sign convention（正方向约定）。抛体与相对速度题先画带坐标的示意图再算。理解收尾速度 $v_{\text{terminal}}$ 存在的原因（阻力与重力平衡），以及有阻力时 $v$-$t$ 图的形状。
Distance vs Displacement, Speed vs Velocity距离与位移、速率与速度 A.1 SL+HL
- Distance $d$ (scalar, path length, always $\ge 0$) vs displacement $\vec{s}$ (vector, straight-line change in position).
- Speed $v$ (scalar, $|\vec{v}|$) vs velocity $\vec{v}$ (vector, rate of change of displacement).
- Acceleration $\vec{a}$ (vector, rate of change of velocity). Units $\mathrm{m\,s^{-2}}$.
- 路程（distance）$d$（标量，路径长度，$\ge 0$）与位移（displacement）$\vec{s}$（矢量，位置的直线变化）。
- 速率（speed）$v$（标量，$|\vec{v}|$）与速度（velocity）$\vec{v}$（矢量，位移变化率）。
- 加速度（acceleration）$\vec{a}$（矢量，速度变化率）。单位 $\mathrm{m\,s^{-2}}$。
A runner jogs $100\ \mathrm{m}$ east, then $40\ \mathrm{m}$ west, in a total time of $30\ \mathrm{s}$. Find the distance travelled, the displacement, the average speed, and the average velocity.跑步者向东 $100\ \mathrm{m}$，再向西 $40\ \mathrm{m}$，总用时 $30\ \mathrm{s}$。求路程、位移、平均速率与平均速度。
Identify. Take east as positive.
识别。取东为正方向。
Distance. Path length: $d = 100 + 40 = 140\ \mathrm{m}$.
路程。路径长度：$d = 100 + 40 = 140\ \mathrm{m}$。
Displacement. Straight-line change in position: $s = (+100) + (-40) = +60\ \mathrm{m}$, i.e. $60\ \mathrm{m}$ east.
位移。位置的直线变化：$s = (+100) + (-40) = +60\ \mathrm{m}$，即向东 $60\ \mathrm{m}$。
Average speed. $v_{\text{avg}} = d / t = 140 / 30 \approx 4.67\ \mathrm{m\,s^{-1}}$.
平均速率。$v_{\text{avg}} = d / t = 140 / 30 \approx 4.67\ \mathrm{m\,s^{-1}}$。
Average velocity. $\vec{v}_{\text{avg}} = \Delta \vec{s} / \Delta t = 60 / 30 = 2.0\ \mathrm{m\,s^{-1}}$ east.
平均速度。$\vec{v}_{\text{avg}} = \Delta \vec{s} / \Delta t = 60 / 30 = 2.0\ \mathrm{m\,s^{-1}}$，向东。
Evaluate. Average speed and average velocity differ in both magnitude and meaning. The runner backtracked, so the path length exceeds the net displacement.
评估。平均速率与平均速度在大小与含义上都不同。跑者有折返，故路径长度大于净位移。
Going deeper: instantaneous from average via a limit深入：由平均量取极限得到瞬时量
Instantaneous velocity is the limit of average velocity as $\Delta t \to 0$:
瞬时速度是平均速度在 $\Delta t \to 0$ 下的极限：
$$ \vec{v}(t) = \lim_{\Delta t \to 0} \frac{\vec{s}(t + \Delta t) - \vec{s}(t)}{\Delta t} = \frac{d \vec{s}}{d t}. $$The same construction yields instantaneous acceleration $\vec{a} = d \vec{v} / d t$. On a position-time graph, this is the tangent slope at the instant of interest.
同理得瞬时加速度 $\vec{a} = d \vec{v} / d t$。在位置-时间图上，对应感兴趣时刻的切线斜率。
The Five suvat Equations五个 suvat 方程 A.1 SL+HL
From the data booklet. $$ v = u + a t, \qquad s = u t + \tfrac{1}{2} a t^{2}, \qquad v^{2} = u^{2} + 2 a s. $$ Two more derived from the above. $$ s = \tfrac{1}{2}(u + v) t, \qquad s = v t - \tfrac{1}{2} a t^{2}. $$ Each suvat connects four of $\{s, u, v, a, t\}$ and omits one. Pick the equation that contains the four quantities the question gives you and the unknown you want.
数据手册公式。 $$ v = u + a t, \qquad s = u t + \tfrac{1}{2} a t^{2}, \qquad v^{2} = u^{2} + 2 a s. $$ 由上面两式推得两个常用式。 $$ s = \tfrac{1}{2}(u + v) t, \qquad s = v t - \tfrac{1}{2} a t^{2}. $$ 每个 suvat 连接 $\{s, u, v, a, t\}$ 中的四个，缺一个。挑选包含题目给出的四个量与所求未知量的那个式子。
| Equation方程 | Missing variable缺哪个量 |
|---|---|
| $v = u + a t$ | $s$ |
| $s = u t + \tfrac{1}{2} a t^{2}$ | $v$ |
| $v^{2} = u^{2} + 2 a s$ | $t$ |
| $s = \tfrac{1}{2}(u + v) t$ | $a$ |
| $s = v t - \tfrac{1}{2} a t^{2}$ | $u$ |
A car accelerates from rest at $2.5\ \mathrm{m\,s^{-2}}$ for $8.0\ \mathrm{s}$. Find (a) the final velocity, (b) the distance travelled.汽车从静止以 $2.5\ \mathrm{m\,s^{-2}}$ 加速 $8.0\ \mathrm{s}$。求 (a) 末速度，(b) 行驶距离。
Identify. Known: $u = 0$, $a = 2.5\ \mathrm{m\,s^{-2}}$, $t = 8.0\ \mathrm{s}$.
识别。已知：$u = 0$、$a = 2.5\ \mathrm{m\,s^{-2}}$、$t = 8.0\ \mathrm{s}$。
(a) Set up. For $v$ with $u, a, t$ known, use $v = u + a t$.
(a) 列式。已知 $u, a, t$ 求 $v$，用 $v = u + a t$。
$$ v = 0 + (2.5)(8.0) = 20\ \mathrm{m\,s^{-1}}. $$(b) Set up. For $s$ with $u, a, t$ known, use $s = u t + \tfrac{1}{2} a t^{2}$.
(b) 列式。已知 $u, a, t$ 求 $s$，用 $s = u t + \tfrac{1}{2} a t^{2}$。
$$ s = 0 + \tfrac{1}{2}(2.5)(8.0)^{2} = 80\ \mathrm{m}. $$Evaluate. Cross-check with the mean-velocity form: $s = \tfrac{1}{2}(u + v) t = \tfrac{1}{2}(0 + 20)(8.0) = 80\ \mathrm{m}$. Consistent.
评估。用平均速度式互校：$s = \tfrac{1}{2}(u + v) t = \tfrac{1}{2}(0 + 20)(8.0) = 80\ \mathrm{m}$，一致。
Going deeper: deriving $v^{2} = u^{2} + 2as$ from the first two深入：由前两式推 $v^{2} = u^{2} + 2as$
From $v = u + a t$, solve for $t = (v - u) / a$. Substitute into $s = u t + \tfrac{1}{2} a t^{2}$:
由 $v = u + a t$ 解出 $t = (v - u) / a$。代入 $s = u t + \tfrac{1}{2} a t^{2}$：
$$ s \;=\; u \cdot \frac{v - u}{a} \;+\; \tfrac{1}{2} a \left(\frac{v - u}{a}\right)^{2} \;=\; \frac{u v - u^{2}}{a} \;+\; \frac{(v - u)^{2}}{2 a}. $$Combine over a common denominator $2 a$:
通分（公分母 $2 a$）：
$$ 2 a s = 2 u v - 2 u^{2} + (v - u)^{2} = 2 u v - 2 u^{2} + v^{2} - 2 u v + u^{2} = v^{2} - u^{2}. $$Therefore $v^{2} = u^{2} + 2 a s$. The equation requires constant $a$.
故 $v^{2} = u^{2} + 2 a s$。该式要求 $a$ 为常量。
Graphical Analysis of Motion运动的图像分析 A.1 SL+HL
- Slope rule. The slope (gradient) of an $x$-$t$ graph is velocity. The slope of a $v$-$t$ graph is acceleration.
- Area rule. The signed area under a $v$-$t$ graph between $t = t_{1}$ and $t = t_{2}$ is the displacement over that interval. The signed area under an $a$-$t$ graph is the change in velocity.
- $x$-$t$: parabola (since $x = u t + \tfrac{1}{2} a t^{2}$).
- $v$-$t$: straight line, slope $a$, intercept $u$.
- $a$-$t$: horizontal line at $y = a$.
- 斜率规则。$x$-$t$ 图斜率即速度。$v$-$t$ 图斜率即加速度。
- 面积规则。$v$-$t$ 图在 $t = t_{1}$ 至 $t = t_{2}$ 之间的有符号面积即该段位移。$a$-$t$ 图的有符号面积即速度变化。
- $x$-$t$：抛物线（$x = u t + \tfrac{1}{2} a t^{2}$）。
- $v$-$t$：直线，斜率 $a$，截距 $u$。
- $a$-$t$：水平线 $y = a$。
A cyclist's velocity-time graph is a trapezium: $v$ rises linearly from $0$ to $8.0\ \mathrm{m\,s^{-1}}$ over $0 \le t \le 4\ \mathrm{s}$, stays at $8.0\ \mathrm{m\,s^{-1}}$ for $4 \le t \le 10\ \mathrm{s}$, then falls linearly to $0$ at $t = 14\ \mathrm{s}$. Find the total displacement and the acceleration in the first interval.骑行者的速度-时间图为梯形：$0 \le t \le 4\ \mathrm{s}$ 时 $v$ 由 $0$ 线性升至 $8.0\ \mathrm{m\,s^{-1}}$；$4 \le t \le 10\ \mathrm{s}$ 保持 $8.0\ \mathrm{m\,s^{-1}}$；之后线性降至 $t = 14\ \mathrm{s}$ 的 $0$。求总位移与第一段加速度。
Identify. The graph is a trapezium with parallel sides $b_{1} = (10 - 4) = 6\ \mathrm{s}$ (top) and $b_{2} = 14\ \mathrm{s}$ (bottom), and height $h = 8.0\ \mathrm{m\,s^{-1}}$.
识别。梯形上底 $b_{1} = 6\ \mathrm{s}$，下底 $b_{2} = 14\ \mathrm{s}$，高 $h = 8.0\ \mathrm{m\,s^{-1}}$。
Total displacement. Area of the trapezium:
总位移。梯形面积：
$$ s = \tfrac{1}{2}(b_{1} + b_{2}) h = \tfrac{1}{2}(6 + 14)(8.0) = 80\ \mathrm{m}. $$Acceleration (first interval). Slope of the rising line: $a = \Delta v / \Delta t = 8.0 / 4 = 2.0\ \mathrm{m\,s^{-2}}$.
加速度（第一段）。上升直线的斜率：$a = \Delta v / \Delta t = 8.0 / 4 = 2.0\ \mathrm{m\,s^{-2}}$。
Evaluate. The trapezium-area shortcut is the same as integrating piecewise: triangle ($16\ \mathrm{m}$) + rectangle ($48\ \mathrm{m}$) + triangle ($16\ \mathrm{m}$) $= 80\ \mathrm{m}$.
评估。梯形面积捷径与分段积分一致：三角形 $16\ \mathrm{m}$ + 矩形 $48\ \mathrm{m}$ + 三角形 $16\ \mathrm{m}$ $= 80\ \mathrm{m}$。
Going deeper: area as the integral $\displaystyle s = \int v\, dt$深入：面积即积分 $\displaystyle s = \int v\, dt$
For any motion (constant or non-constant acceleration), the displacement between $t_{1}$ and $t_{2}$ is
对任意运动（匀加速或变加速），$t_{1}$ 到 $t_{2}$ 间的位移为
$$ \Delta s = \int_{t_{1}}^{t_{2}} v(t)\, dt. $$Geometrically this is the signed area under the $v(t)$ curve. For constant $a$, the line is straight and the integral becomes the trapezium-area formula. For drag-affected motion (A1.6), $v(t)$ curves and you would estimate the area by counting grid squares or by Simpson's rule.
几何上即 $v(t)$ 曲线下的有符号面积。匀加速时直线下的积分就是梯形面积公式。带阻力的运动（A1.6）中 $v(t)$ 是曲线，需要数格子或用 Simpson 法估面积。
Free Fall and Projectile Motion自由落体与抛体运动 A.1 SL+HL
Projectile motion (2D, no drag). The horizontal and vertical components are independent: $$ a_{x} = 0, \qquad a_{y} = -g. $$ Decomposition of an initial velocity $v_{0}$ at angle $\theta$ above horizontal. $$ v_{0x} = v_{0} \cos\theta, \qquad v_{0y} = v_{0} \sin\theta. $$ Position equations. $$ x(t) = v_{0x}\, t, \qquad y(t) = v_{0y}\, t - \tfrac{1}{2} g t^{2}. $$ Useful derived results (level ground, launch height $= 0$). $$ T_{\text{flight}} = \frac{2 v_{0} \sin\theta}{g}, \qquad R = \frac{v_{0}^{2} \sin 2\theta}{g}, \qquad H = \frac{v_{0}^{2} \sin^{2}\theta}{2 g}. $$ Maximum range on level ground: $\theta = 45^{\circ}$.
抛体运动（二维，无阻力）。水平与竖直分量相互独立： $$ a_{x} = 0, \qquad a_{y} = -g. $$ 初速度 $v_{0}$ 与水平方向夹角 $\theta$ 时的分量分解。 $$ v_{0x} = v_{0} \cos\theta, \qquad v_{0y} = v_{0} \sin\theta. $$ 位置方程。 $$ x(t) = v_{0x}\, t, \qquad y(t) = v_{0y}\, t - \tfrac{1}{2} g t^{2}. $$ 等高地面（发射高度 $= 0$）常用推论。 $$ T_{\text{flight}} = \frac{2 v_{0} \sin\theta}{g}, \qquad R = \frac{v_{0}^{2} \sin 2\theta}{g}, \qquad H = \frac{v_{0}^{2} \sin^{2}\theta}{2 g}. $$ 等高时最大射程对应 $\theta = 45^{\circ}$。
A ball is thrown horizontally at $15\ \mathrm{m\,s^{-1}}$ from a cliff $20\ \mathrm{m}$ above level ground. Find the time of flight and the horizontal range.小球从离地 $20\ \mathrm{m}$ 的悬崖以 $15\ \mathrm{m\,s^{-1}}$ 水平抛出。求飞行时间与水平射程。
Set up. Take down as positive. Vertically: $u_{y} = 0$, $a_{y} = g = 9.81\ \mathrm{m\,s^{-2}}$, $s_{y} = 20\ \mathrm{m}$. Horizontally: $u_{x} = 15\ \mathrm{m\,s^{-1}}$, $a_{x} = 0$.
列式。取向下为正。竖直方向：$u_{y} = 0$、$a_{y} = g = 9.81\ \mathrm{m\,s^{-2}}$、$s_{y} = 20\ \mathrm{m}$。水平方向：$u_{x} = 15\ \mathrm{m\,s^{-1}}$、$a_{x} = 0$。
Time of flight. Use $s_{y} = u_{y} t + \tfrac{1}{2} g t^{2}$: $20 = \tfrac{1}{2}(9.81) t^{2} \Rightarrow t = \sqrt{40/9.81} \approx 2.02\ \mathrm{s}$.
飞行时间。用 $s_{y} = u_{y} t + \tfrac{1}{2} g t^{2}$：$20 = \tfrac{1}{2}(9.81) t^{2} \Rightarrow t = \sqrt{40/9.81} \approx 2.02\ \mathrm{s}$。
Horizontal range. $x = u_{x} t = (15)(2.02) \approx 30.3\ \mathrm{m}$.
水平射程。$x = u_{x} t = (15)(2.02) \approx 30.3\ \mathrm{m}$。
Evaluate. The horizontal launch has no vertical component, so $t$ depends only on the fall height. Once $t$ is known, multiply by $u_{x}$ for the range.
评估。水平抛出无竖直初速度，故 $t$ 仅由下落高度决定；再乘 $u_{x}$ 得射程。
A projectile is launched at $25\ \mathrm{m\,s^{-1}}$ at $30^{\circ}$ above the horizontal from ground level. Find the time of flight, range, and maximum height. Take $g = 9.81\ \mathrm{m\,s^{-2}}$.弹丸以 $25\ \mathrm{m\,s^{-1}}$、与水平方向夹角 $30^{\circ}$ 自地面发射。求飞行时间、射程与最大高度。取 $g = 9.81\ \mathrm{m\,s^{-2}}$。
Decompose. $v_{0x} = 25 \cos 30^{\circ} \approx 21.65\ \mathrm{m\,s^{-1}}$, $v_{0y} = 25 \sin 30^{\circ} = 12.5\ \mathrm{m\,s^{-1}}$.
分解。$v_{0x} = 25 \cos 30^{\circ} \approx 21.65\ \mathrm{m\,s^{-1}}$、$v_{0y} = 25 \sin 30^{\circ} = 12.5\ \mathrm{m\,s^{-1}}$。
Time of flight (return to $y = 0$).
飞行时间（返回 $y = 0$）。
$$ T = \frac{2 v_{0y}}{g} = \frac{2 (12.5)}{9.81} \approx 2.55\ \mathrm{s}. $$Range.
射程。
$$ R = v_{0x}\, T = (21.65)(2.55) \approx 55.2\ \mathrm{m}. $$Maximum height. At the apex, $v_{y} = 0$, so use $v_{y}^{2} = v_{0y}^{2} - 2 g H$:
最大高度。在最高点 $v_{y} = 0$，用 $v_{y}^{2} = v_{0y}^{2} - 2 g H$：
$$ H = \frac{v_{0y}^{2}}{2 g} = \frac{(12.5)^{2}}{2 (9.81)} \approx 7.96\ \mathrm{m}. $$Evaluate. Independence check: $R$ matches the data-booklet form $R = v_{0}^{2} \sin 2\theta / g = 625 \sin 60^{\circ} / 9.81 \approx 55.2\ \mathrm{m}$.
评估。独立性检验：$R$ 与数据手册公式 $R = v_{0}^{2} \sin 2\theta / g = 625 \sin 60^{\circ} / 9.81 \approx 55.2\ \mathrm{m}$ 一致。
Going deeper: why $45^{\circ}$ maximises range on level ground深入：等高地面为何 $45^{\circ}$ 最大化射程
From the range formula $R = v_{0}^{2} \sin 2\theta / g$, treat $v_{0}$ and $g$ as fixed. The factor $\sin 2\theta$ is maximised when $2\theta = 90^{\circ}$, i.e. $\theta = 45^{\circ}$, giving $\sin 2\theta = 1$ and $R_{\max} = v_{0}^{2} / g$.
由射程公式 $R = v_{0}^{2} \sin 2\theta / g$，$v_{0}$、$g$ 固定时仅 $\sin 2\theta$ 可变；$\sin 2\theta$ 在 $2\theta = 90^{\circ}$（即 $\theta = 45^{\circ}$）取最大值 $1$，故 $R_{\max} = v_{0}^{2} / g$。
Two angles equidistant from $45^{\circ}$ (for example $30^{\circ}$ and $60^{\circ}$) give the same range, because $\sin 2\theta$ is symmetric about $\theta = 45^{\circ}$. This is a useful sanity check.
关于 $45^{\circ}$ 对称的两角（如 $30^{\circ}$ 与 $60^{\circ}$）给出相同射程，因为 $\sin 2\theta$ 关于 $\theta = 45^{\circ}$ 对称。常用于复核答案。
Caveat. If the launch and landing heights differ (e.g. launching off a cliff), the optimum angle is less than $45^{\circ}$; the data-booklet $R$ formula no longer applies and you must solve from $y(t) = 0$ directly.
注意。若发射与落地不等高（如从崖上发射），最优角度小于 $45^{\circ}$；数据手册的 $R$ 公式不再适用，需直接由 $y(t) = 0$ 解出。
2D Kinematics: Components and Relative Velocity二维运动学：分量与相对速度 A.1 SL+HL
Common patterns.
- Two cars approaching head-on at $u$ and $v$ have relative speed $u + v$.
- Two cars moving in the same direction at $u$ and $v$ have relative speed $|u - v|$.
- For 2D (a boat in a current, plane in wind), draw the vector triangle and use Pythagoras / trigonometry.
常见情形。
- 两车迎面而来，速率分别为 $u$、$v$，相对速率 $u + v$。
- 同向行驶，速率分别为 $u$、$v$，相对速率 $|u - v|$。
- 二维情形（船在水流中、飞机在风中）画矢量三角形，用勾股或三角函数求解。
A boat travels at $4.0\ \mathrm{m\,s^{-1}}$ relative to the water, aimed directly north across a river that flows east at $3.0\ \mathrm{m\,s^{-1}}$ relative to the bank. Find the boat's velocity relative to the bank (magnitude and direction). If the river is $120\ \mathrm{m}$ wide, how far downstream does the boat land?船相对水流速度 $4.0\ \mathrm{m\,s^{-1}}$，船头直指正北；河水相对河岸向东流速 $3.0\ \mathrm{m\,s^{-1}}$。求船相对河岸的速度（大小与方向）。若河宽 $120\ \mathrm{m}$，船被冲下游多少？
Set up. Take east as $+x$, north as $+y$. Velocity of boat relative to water $\vec{v}_{BW} = (0, 4.0)$. Velocity of water relative to bank $\vec{v}_{WG} = (3.0, 0)$. Velocity of boat relative to bank:
列式。取东为 $+x$、北为 $+y$。船相对水 $\vec{v}_{BW} = (0, 4.0)$；水相对岸 $\vec{v}_{WG} = (3.0, 0)$。船相对岸：
$$ \vec{v}_{BG} = \vec{v}_{BW} + \vec{v}_{WG} = (3.0,\ 4.0)\ \mathrm{m\,s^{-1}}. $$Magnitude and direction.
大小与方向。
$$ |\vec{v}_{BG}| = \sqrt{3.0^{2} + 4.0^{2}} = 5.0\ \mathrm{m\,s^{-1}}, \qquad \theta = \arctan(3.0 / 4.0) \approx 36.9^{\circ}\ \text{east of north}. $$Time to cross. Northward distance: $120\ \mathrm{m}$; northward velocity: $4.0\ \mathrm{m\,s^{-1}}$.
过河时间。北向距离 $120\ \mathrm{m}$，北向速度 $4.0\ \mathrm{m\,s^{-1}}$。
$$ t = 120 / 4.0 = 30\ \mathrm{s}. $$Downstream drift. $\Delta x = v_{x}\, t = (3.0)(30) = 90\ \mathrm{m}$.
下游漂移。$\Delta x = v_{x}\, t = (3.0)(30) = 90\ \mathrm{m}$。
Evaluate. Crossing time is set entirely by the perpendicular (north) component; the current does not change how long the boat is on the water, only where it lands.
评估。过河时间仅由垂直（北向）分量决定；水流不影响过河时间，只影响落点。
Going deeper: relative velocity in a moving reference frame HL only深入：动参考系中的相对速度 HL only
If $S$ is a "ground" frame and $S'$ moves at constant velocity $\vec{V}$ relative to $S$, an object whose velocity in $S$ is $\vec{u}$ has velocity in $S'$ given by
若 $S$ 为"地面"参考系，$S'$ 以恒速 $\vec{V}$ 相对 $S$ 运动，则 $S$ 系中速度为 $\vec{u}$ 的物体在 $S'$ 系中速度为
$$ \vec{u}' = \vec{u} - \vec{V}. $$This is the Galilean velocity addition rule. It is exact at speeds much less than $c$ and is the result you use throughout A.1. The exact rule that applies at relativistic speeds is taken up in unit A.5.
这是伽利略速度合成法则。当速率远小于 $c$ 时严格成立，A.1 单元自始至终用它。相对论修正在单元 A.5 中讨论。
Two consequences:
两个推论：
- Accelerations are frame-invariant under Galilean boosts: $\vec{a}' = \vec{a}$. Newton's second law has the same form in every inertial frame.
- 伽利略变换下加速度不变：$\vec{a}' = \vec{a}$。牛顿第二定律在每个惯性系中形式相同。
- A ball thrown straight up inside a uniformly-moving train falls back into the thrower's hand, because the horizontal motion of train and ball is identical in the train frame.
- 在匀速行驶的火车内竖直上抛的小球会落回抛者手中；因为在火车参考系内火车与球的水平运动相同。
Fluid Resistance and Terminal Velocity流体阻力与收尾速度 HL only A.1 AHL
- Low speed (laminar regime): $F_{D} = k v$ (linear in speed).
- High speed (turbulent regime): $F_{D} \propto v^{2}$ (quadratic in speed).
Effect on motion. Acceleration is no longer constant, so the suvat equations fail. The $v$-$t$ graph of a falling object with drag rises with decreasing slope and asymptotically approaches $v_{T}$.
- 低速（层流）：$F_{D} = k v$，与速率成正比。
- 高速（湍流）：$F_{D} \propto v^{2}$，与速率平方成正比。
对运动的影响。加速度不再恒定，suvat 方程失效。带阻力下落的 $v$-$t$ 图斜率递减，渐近趋向 $v_{T}$。
A small bead of mass $0.020\ \mathrm{kg}$ falls through oil that exerts a linear drag $F_{D} = k v$ with $k = 0.040\ \mathrm{N\,s\,m^{-1}}$. Find the terminal velocity (take $g = 9.81\ \mathrm{m\,s^{-2}}$) and describe the $v$-$t$ graph.质量 $0.020\ \mathrm{kg}$ 的小珠在油中下落，所受线性阻力 $F_{D} = k v$，$k = 0.040\ \mathrm{N\,s\,m^{-1}}$。求收尾速度（取 $g = 9.81\ \mathrm{m\,s^{-2}}$）并描述 $v$-$t$ 图。
Set up. Take down as positive. Forces on the bead: weight $m g$ downward, drag $k v$ upward. Newton's second law: $m \dfrac{d v}{d t} = m g - k v$.
列式。取向下为正。受力：重力 $m g$ 向下，阻力 $k v$ 向上。牛顿第二定律：$m \dfrac{d v}{d t} = m g - k v$。
Terminal condition. $d v / d t = 0$ gives $m g = k v_{T}$.
收尾条件。$d v / d t = 0$，得 $m g = k v_{T}$。
$$ v_{T} = \frac{m g}{k} = \frac{(0.020)(9.81)}{0.040} \approx 4.91\ \mathrm{m\,s^{-1}}. $$$v$-$t$ shape. Initially $v = 0$, so drag is zero and acceleration equals $g$ (slope of the $v$-$t$ graph is $g$ at $t = 0$). As $v$ increases, the slope decreases, asymptotically reaching $v_{T} \approx 4.91\ \mathrm{m\,s^{-1}}$ but never exceeding it.
$v$-$t$ 形状。初始 $v = 0$，阻力为零，加速度等于 $g$（$t = 0$ 时 $v$-$t$ 图斜率为 $g$）。随 $v$ 增大，斜率减小，渐近趋向 $v_{T} \approx 4.91\ \mathrm{m\,s^{-1}}$，但不会超过它。
Evaluate. The bead never quite reaches $v_{T}$ but for practical purposes is within a few percent after a characteristic time $\tau = m / k = 0.020 / 0.040 = 0.5\ \mathrm{s}$.
评估。小珠不会真正达到 $v_{T}$，但经过特征时间 $\tau = m / k = 0.5\ \mathrm{s}$ 后已接近至几个百分点内。
Going deeper: solving $m \dot v = m g - k v$深入：求解 $m \dot v = m g - k v$
The equation $m \dfrac{d v}{d t} = m g - k v$ is first-order linear. Rewriting with $v_{T} = m g / k$:
方程 $m \dfrac{d v}{d t} = m g - k v$ 为一阶线性方程。用 $v_{T} = m g / k$ 改写：
$$ \frac{d v}{d t} = \frac{g}{v_{T}} (v_{T} - v) = \frac{1}{\tau} (v_{T} - v), \qquad \tau = m / k. $$Separating variables and integrating with $v(0) = 0$:
分离变量并以初值 $v(0) = 0$ 积分：
$$ v(t) = v_{T} \left( 1 - e^{-t / \tau} \right). $$This is an exponential approach to $v_{T}$ with time constant $\tau$. At $t = \tau$, $v$ has reached $63\%$ of $v_{T}$; at $t = 3 \tau$, $95\%$; at $t = 5 \tau$, $99.3\%$. The IB syllabus does not require deriving this explicitly, but recognising the exponential shape on a sketched $v$-$t$ graph is expected.
即以时间常数 $\tau$ 指数式趋向 $v_{T}$。$t = \tau$ 时 $v$ 达 $v_{T}$ 的 $63\%$；$t = 3\tau$ 时 $95\%$；$t = 5\tau$ 时 $99.3\%$。IB 大纲不要求显式推导，但要求识别 $v$-$t$ 图的指数形态。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- State which direction you choose as positive at the start of your working. Markschemes credit consistent signs; mixed signs lose A1 marks.
- 在解题开头写明所选正方向。评分要求符号一致；混用符号会丢 A1。
- For free fall, "up positive" gives $a = -g$; "down positive" gives $a = +g$. Pick one and stick to it.
- 自由落体中"向上为正"对应 $a = -g$；"向下为正"对应 $a = +g$。选定后不要中途切换。
- List $s, u, v, a, t$ before reaching for a formula. The missing variable points at the equation.
- 动笔前先列出 $s, u, v, a, t$ 五个量。缺哪个就对应哪条公式。
- If no time is given, use $v^{2} = u^{2} + 2 a s$. This is the most-asked decelerate-to-stop and projectile-height variant.
- 若未给时间，用 $v^{2} = u^{2} + 2 a s$。"减速到停"与"求抛体最高高度"几乎都用它。
- Decompose the initial velocity once at the top of your working. Then treat $x$ and $y$ as independent 1D problems.
- 解题开头把初速度一次性分解为 $v_{0x}$ 与 $v_{0y}$。之后把 $x$、$y$ 当作两个独立的一维问题处理。
- Use $T = 2 v_{0y} / g$ only on level ground. If launch height differs from landing height, solve $y(t) = h_{\text{landing}}$ directly.
- $T = 2 v_{0y} / g$ 仅在等高地面适用。发射与落地不等高时，直接解 $y(t) = h_{\text{landing}}$。
- "Find the velocity at $t = T$" on an $x$-$t$ graph means "draw the tangent and measure its slope". State both rise/run figures.
- "在 $x$-$t$ 图上求 $t = T$ 时的速度"即"画切线、量斜率"。同时给出"上升"与"水平"两个读数。
- "Find the displacement between $t_{1}$ and $t_{2}$" on a $v$-$t$ graph means "compute the signed area". Use trapezia or grid-square counting.
- "在 $v$-$t$ 图上求 $t_{1}$ 与 $t_{2}$ 间的位移"即"求有符号面积"。用梯形或数格子。
Flashcards闪卡
$$v_{0y} = v_{0} \sin\theta$$
Unit A.1 Practice Quiz单元 A.1 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- Distinguish distance from displacement and speed from velocity, with signed answers in 1D在一维带符号情境中区分路程与位移、速率与速度
- State a sign convention before setting up any motion problem在列式前写明正方向约定
- List $s, u, v, a, t$ from the question and choose the correct suvat equation从题目中列出 $s, u, v, a, t$ 并选对 suvat 方程
- Compute final velocity, displacement, or acceleration with constant-$a$ motion in 1D在一维匀加速运动中求末速度、位移或加速度
- Read velocity from the slope of an $x$-$t$ graph by drawing a tangent通过切线读出 $x$-$t$ 图的速度
- Compute displacement from the signed area of a $v$-$t$ graph (trapezia or grid counting)由 $v$-$t$ 图的有符号面积求位移（梯形或数格子）
- Solve a 1D free-fall problem (drop, throw upward, throw downward) with $g = 9.81\ \mathrm{m\,s^{-2}}$用 $g = 9.81\ \mathrm{m\,s^{-2}}$ 求一维自由落体（释放、上抛、下抛）问题
- Decompose an initial velocity into $v_{0x}$ and $v_{0y}$ and solve a projectile problem把初速度分解为 $v_{0x}, v_{0y}$ 后求解抛体问题
- Use the level-ground formulas $T = 2 v_{0} \sin\theta / g$ and $R = v_{0}^{2} \sin 2\theta / g$ correctly正确使用等高地面公式 $T = 2 v_{0} \sin\theta / g$ 与 $R = v_{0}^{2} \sin 2\theta / g$
- Apply $\vec{v}_{AB} = \vec{v}_{A} - \vec{v}_{B}$ to 1D and 2D relative-velocity problems对一维与二维相对速度问题用 $\vec{v}_{AB} = \vec{v}_{A} - \vec{v}_{B}$
- HL Find terminal velocity from the drag-weight balance for a given drag law由阻力与重力平衡求给定阻力律下的收尾速度
- HL Sketch a $v$-$t$ graph for fall with linear drag, showing initial slope $g$ and asymptote $v_{T}$画出线性阻力下落的 $v$-$t$ 图：初始斜率 $g$，渐近线 $v_{T}$
IB Paper-Style PracticeIB 试卷风格练习
A.1 Practice and Solutions are on the roadmap, to ship under Practice Questions/Unit_A1_*.html with the bilingual built-in pattern.
A.1 配套 Practice 与 Solutions 在排期，上线后位于 Practice Questions/Unit_A1_*.html。