Describe and predict the motion of objects using vectors, calculus-based definitions of velocity and acceleration, kinematic equations, and the independence of perpendicular motion components.用矢量（vector）、基于微积分的速度（velocity）与加速度（acceleration）定义、运动学方程（kinematic equations），以及垂直方向上运动分量相互独立的原理，描述并预测物体的运动。
10–15% Exam Weight考试占分 10–15%~14–19 Class Periods约 14–19 课时5 Topics5 个专题
Topic 1.1专题 1.1
Scalars and Vectors标量与矢量
Physics begins with measurement, and the first distinction to make is between quantities that have direction and those that don't. A scalar is described by magnitude alone — distance, speed, mass, and energy are all scalars. A vector requires both magnitude and direction — position, displacement, velocity, acceleration, and force are vectors.
Visual Model
Vectors are drawn as arrows. The arrow's length is proportional to the magnitude, and its orientation shows the direction. In a given one-dimensional coordinate system, opposite directions are denoted by opposite signs.
Vectors can be expressed in two equivalent ways. As a magnitude and angle ($|\vec{A}|$ at $\theta$), or in unit vector notation using $\hat{i}$, $\hat{j}$, $\hat{k}$ for the $x$-, $y$-, $z$-directions respectively. The position vector of a point is $\vec{r}$, and the unit vector in its direction is $\hat{r}$.
Exam Tip
Always decompose vectors into components before calculating — never try to add vectors by "tip-to-tail" numerically. On the AP exam, remember that $\arctan$ gives you an angle relative to the $x$-axis, and you must manually adjust for quadrants II and III.
When we model a physical object, we often use the object (particle) model — ignoring its size, shape, and internal structure and treating it as a single point. This simplification lets us focus entirely on the object's position as a function of time.
Distance vs. DisplacementDistance (scalar): total path length traveled — always positive.Displacement (vector): change in position — has direction and can be negative.A round trip has zero displacement but nonzero distance.
距离 vs. 位移距离（distance，标量）：经过的总路径长度 —— 永远为正。位移（displacement，矢量）：位置的变化 —— 有方向，可正可负。往返一周后位移为零，但距离非零。
Displacement位移
$$\Delta x = x - x_0$$
Average vs. Instantaneous平均量与瞬时量
Average quantities are computed over a finite time interval using initial and final states. As the time interval shrinks toward zero, the average approaches the instantaneous value — and this limit is exactly the derivative.
Acceleration is the time derivative of velocity (second derivative of position).
加速度是速度对时间的导数（也就是位置对时间的二阶导数）。
Common Misconception
An object is accelerating whenever its velocity changes — in magnitude and/or direction. Don't confuse "negative acceleration" with "slowing down." Negative acceleration simply means the acceleration vector points in the negative direction. The object slows down only when $v$ and $a$ have opposite signs.
Exam Tip — Show Your Calculus
This is AP Physics C. Always use derivatives for instantaneous quantities and integrals for accumulation. Free-response graders award explicit points for writing $v = dx/dt$ and showing the differentiation or integration step. Always include the constant of integration and use initial conditions to determine it.
Motion can be represented through motion diagrams, figures, graphs, equations, and narrative descriptions. The most important skill in kinematics is translating fluently between these representations — this is the core of the Translation Between Representations (TBR) free-response question.
运动可以通过运动示意图、图形、图像（graph）、方程和文字描述来表示。运动学最关键的能力是在这些表示之间自如切换 —— 这正是 AP 考试中"表示之间的转换"（Translation Between Representations / TBR）自由作答题的核心。
When acceleration is constant, three fundamental kinematic equations describe motion in one dimension. These are derived from calculus — they are the results of integrating constant $a$.
The connections between position, velocity, and acceleration graphs are the backbone of kinematics analysis. Slopes give derivatives; areas give integrals.
位置、速度、加速度三种图像（graph）之间的联系是运动学分析的脊梁。斜率（slope）就是导数；曲线下面积（area under the curve）就是积分。
Graph ↔ Calculus Connections图像 ↔ 微积分对应关系
From
To
Operation
On the Graph
$x(t)$
$v(t)$
$v = dx/dt$
Slope of $x$-$t$ graph
$v(t)$
$a(t)$
$a = dv/dt$
Slope of $v$-$t$ graph
$v(t)$
$\Delta x$
$\Delta x = \int v\,dt$
Area under $v$-$t$ curve
$a(t)$
$\Delta v$
$\Delta v = \int a\,dt$
Area under $a$-$t$ curve
从
到
运算
在图像上
$x(t)$
$v(t)$
$v = dx/dt$
$x$-$t$ 图的斜率
$v(t)$
$a(t)$
$a = dv/dt$
$v$-$t$ 图的斜率
$v(t)$
$\Delta x$
$\Delta x = \int v\,dt$
$v$-$t$ 曲线下面积
$a(t)$
$\Delta v$
$\Delta v = \int a\,dt$
$a$-$t$ 曲线下面积
Displacement from Velocity由速度求位移
$$\Delta x = \int_{t_1}^{t_2} v_x(t)\;dt$$
Velocity Change from Acceleration由加速度求速度变化
$$\Delta v_x = \int_{t_1}^{t_2} a_x(t)\;dt$$
Free Fall自由落体
Gravitational Acceleration
Near Earth's surface, the vertical acceleration due to gravity is downward, constant, and approximately $g \approx 10\;\text{m/s}^2$. On the AP exam, always use $g \approx 10$ m/s² unless otherwise directed.
The ball is thrown straight up — its motion is purely vertical, with no horizontal component. The left strip shows the ball's actual stroboscopic positions on a single vertical line. The right panel is a height-vs-time graph (time on the horizontal axis, NOT horizontal distance). The parabola shape is the function $y(t)$, not the path through space.
The dots in the left strip show where the ball physically is at $t = 0, 0.5, 1.0, \ldots$ s — they sit on a single vertical line because the motion is purely 1D. Purple dots are on the way up, green is the apex, red is on the way down. The same data plotted as a graph (right) gives the parabola $y(t) = h_0 + v_0 t - \tfrac{1}{2}g t^2$ — that curve is a function plot, not a flight path.
Peak Height
20.0
m
Time to Peak
2.00
s
Total Flight
4.00
s
Impact Speed
20.0
m/s
Exam Trap — Signs
If you choose up as positive, then $a = -g$ (not $+g$). This is the most common source of sign errors on the AP exam. Be consistent with your sign convention and state it explicitly on free-response.
Use $\Delta y = \frac{1}{2}gt^2$: $80 = 5t^2$, so $t^2 = 16$, $t = 4$ s.用公式 $\Delta y = \frac{1}{2}gt^2$：$80 = 5t^2$，得 $t^2 = 16$，$t = 4$ s。
Topic 1.4专题 1.4
Reference Frames and Relative Motion参考系与相对运动
A reference frame defines the origin and coordinate system from which an observer makes measurements. Two observers in different reference frames may measure different velocities for the same object. The key tool for connecting their measurements is the relative velocity equation.
Read: "velocity of A relative to C = velocity of A relative to B + velocity of B relative to C." Inner subscripts cancel.
读作："A 相对 C 的速度 = A 相对 B 的速度 + B 相对 C 的速度"。中间的下标相互抵消。
Galilean Invariance
The acceleration of any object is the same as measured from all inertial reference frames. This means Newton's second law holds equally in every non-accelerating frame. Unless otherwise stated, assume the reference frame is inertial on the AP exam.
Car A: 25 m/s east. Car B: 15 m/s east. Both relative to ground. Velocity of A relative to B?
甲车 25 m/s 向东，乙车 15 m/s 向东（均相对地面）。 甲相对乙的速度是多少？
$$v(A/B) = v(A/G) − v(B/G)$$
$$= 25 − 15$$
$$= 10\;\text{m/s} east$$
Exam Tip — Subscript Notation
The key trick is that inner subscripts cancel: $\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$. Draw a diagram for 2D problems — many students lose points by reversing the direction of the current or wind.
The central insight of multidimensional kinematics is the independence of perpendicular components. Motion in one dimension can be changed without causing a change in a perpendicular dimension. This means 2D or 3D motion is analyzed by separating it into independent 1D kinematic problems, one for each axis.
多维运动学的核心思想是各垂直方向相互独立（independence of perpendicular components）。一个方向上的运动改变并不会引起另一垂直方向上的变化。因此处理二维、三维运动时，把它沿坐标轴拆成几组互不相关的一维运动学问题来分析。
Projectile Motion
Projectile motion is a special case of 2D motion with $a_x = 0$ (constant horizontal velocity) and $a_y = -g$ (constant downward acceleration). Air resistance is neglected. Time is the link between horizontal and vertical motion.
Whenever an FRQ asks "does it clear the wall?" or "what angle must it be launched at to hit target $(x_T, y_T)$?", start from $x(t)$ and $y(t)$, solve $x(t)$ for $t$, and substitute into $y(t)$.
Step 1 — Parametric equations (launch from the origin):
Worked Example — Projectile + Relative Motion FRQ (Plane Drop)例题 —— 抛体 + 相对运动 FRQ（飞机投物）
A plane flies horizontally at $v_p = 80$ m/s at an altitude of $h = 500$ m and releases a supply package. (a) How long until the package lands? (b) What horizontal distance does the package travel relative to the ground? (c) In the plane's reference frame, what does the package's path look like? (d) A target on the ground is moving in the same direction as the plane at $v_t = 20$ m/s. How far behind the target must the package be released so it lands on the target? (Take $g = 10\;\text{m/s}^2$.)
飞机在海拔 $h = 500$ m 处以 $v_p = 80$ m/s 水平飞行，并释放一个补给包。(a) 多久后落地？(b) 相对地面水平移动多远？(c) 在飞机的参考系（plane's reference frame）中包的运动轨迹是什么样？(d) 地面目标与飞机同向、以 $v_t = 20$ m/s 移动。包应在目标后方多远处释放才能正中目标？（取 $g = 10\;\text{m/s}^2$）
(a) Fall time(a) 下落时间
Vertical motion starts with $v_y = 0$ and falls $h = 500$ m:竖直方向初速度 $v_y = 0$，下降 $h = 500$ m：
The package keeps the plane's horizontal velocity at release:释放瞬间包继承飞机的水平速度：
$$x = v_p\,t = (80)(10) = 800\;\text{m}$$
(c) Path in the plane's reference frame(c) 飞机参考系下的轨迹
In the plane's frame the package has zero horizontal velocity at release, so it falls straight down — a vertical line directly below the plane. The plane observer sees pure free fall; the ground observer sees a parabola. Both are correct descriptions of the same motion.在飞机的参考系里，释放瞬间包没有水平速度，因此沿正下方做自由落体（free fall）——是一条紧贴飞机正下方的竖直线。飞机上的观察者看到的是纯粹的自由落体，地面上的观察者看到的是抛物线轨迹。两者描述的是同一运动，都正确。
(d) Release offset for a moving target(d) 对运动目标的释放提前量
Horizontal velocity of the package relative to the target:包相对目标的水平速度：
So the package must be released 600 m behind the target. Ground-frame check: the package travels 800 m while the target drifts only 200 m during the 10 s fall, so release position = target's start $- 600$ m. ✓因此包必须在目标后方 600 m 处释放。用地面参考系核对：在 10 s 内，包水平方向移动 800 m，目标只移动 200 m，所以释放位置 = 目标初位置 $- 600$ m。✓
Frame-Switching Rule
Questions like "what does observer B see?" are just $\vec{v}_{A/B} = \vec{v}_{A/C} - \vec{v}_{B/C}$. For a dropped object in a moving frame, the object inherits the frame's velocity at release. Gravity is the same in all inertial frames.
参考系切换规则
"观察者 B 看到的是什么？"这种问题，只是套用 $\vec{v}_{A/B} = \vec{v}_{A/C} - \vec{v}_{B/C}$。从运动参考系里释放的物体，继承释放瞬间该参考系的速度。重力在所有惯性参考系中均相同。
Projectile Motion Explorer抛体运动交互工具
Adjust launch speed and angle — see range, max height, flight time, and velocity components update live.调节发射速度和角度——实时查看射程、最大高度、飞行时间和速度分量。
Range射程
90.0
m
Max Height最大高度
22.5
m
Flight Time飞行时间
4.24
s
v_x
21.2
m/s (constant)m/s（恒定）
v_y₀
21.2
m/s (initial)m/s（初始）
Common Mistake
At the top of its trajectory, a projectile has $v_y = 0$ but $a_y = -g \neq 0$. Do not confuse zero velocity with zero acceleration. Also, the range formula only works when launch and landing heights are the same — for cliff problems, solve the full kinematic equations separately.
A projectile is launched at 50 m/s at 53° ($\sin 53° = 0.8$, $\cos 53° = 0.6$). What is the maximum height? ($g = 10$ m/s²)抛体以 50 m/s、53° 角发射（$\sin 53° = 0.8$，$\cos 53° = 0.6$）。最大高度是多少？（$g = 10$ m/s²）
How Unit 1 Appears on the AP Exam第 1 单元在 AP 考试中的形式
MC
Multiple Choice — Common Styles
Calculate an unknown kinematic quantity from given information using the three equations or calculus.
Interpret graphs — read slopes and areas from $x$-$t$, $v$-$t$, and $a$-$t$ plots.
Compare scenarios — determine which object arrives first, goes higher, or travels farther.
Relative motion — find velocity of one object as seen from another reference frame.
MC
选择题（Multiple Choice）—— 常见题型
计算类：由已知量用三大运动学方程或微积分求未知量。
图像解读：从 $x$-$t$、$v$-$t$、$a$-$t$ 图中读取斜率与面积。
情景比较：判断哪个物体先到、跳得更高或走得更远。
相对运动：在另一参考系中观察某物体的速度。
FR
Free Response — Common Styles
Translation Between Representations (TBR): Given one representation (e.g., a v-t graph), produce another (e.g., the x-t graph or a motion description).
Experimental design: Describe a procedure to determine kinematic quantities from real data.
Multi-step projectile problems: Derive expressions, calculate unknowns, and explain physical reasoning — all with explicit calculus.
FR
自由回答题（Free Response）—— 常见题型
表示形式之间的转换（Translation Between Representations, TBR）：给一种表示（如 v-t 图），转换为另一种（如 x-t 图或文字描述）。
实验设计：设计一套测量某些运动学量的实际实验方案。
多步抛体题：推导表达式、计算未知量并解释物理意义——通常要明确使用微积分。
Top Mistakes That Lose Points1. Sign errors with $g$ — if up is positive, $a = -g$, not $+g$.2. Applying kinematic equations when acceleration is not constant — use calculus instead.3. Forgetting the constant of integration $C$ when integrating $v(t)$ or $a(t)$.4. Using the range formula for non-level-ground problems.5. Confusing $v = 0$ with $a = 0$ at the peak of a projectile.6. Mixing distance and displacement — a round trip has $\Delta x = 0$ but $d > 0$.7. Not including units in final answers.
$v(t) = 6t^2 - 6t + 1 = 0$. Use the quadratic formula to find the two times when velocity is zero.$v(t) = 6t^2 - 6t + 1 = 0$。用求根公式（quadratic formula）求出两个速度为零的时刻。
2. An object's $v$-$t$ graph shows a straight line from $v = 10$ m/s at $t = 0$ to $v = -10$ m/s at $t = 4$ s. What is the displacement from $t = 0$ to $t = 4$ s?2. 某物体的 $v$-$t$ 图是一条直线，从 $t = 0$ 时 $v = 10$ m/s 到 $t = 4$ s 时 $v = -10$ m/s。$t = 0$ 到 $t = 4$ s 的位移是多少？
$40$ m
$20$ m
$0$ m
$-20$ m
Correct! The area under the $v$-$t$ line is a triangle above the axis (base 2 s, height 10) plus a triangle below (base 2 s, depth −10). Total area = $\frac{1}{2}(2)(10) + \frac{1}{2}(2)(-10) = 10 - 10 = 0$ m.正确！$v$-$t$ 直线下方面积 = 横轴上方三角形（底 2 s，高 10）加上横轴下方三角形（底 2 s，高 -10）。合计 $\frac{1}{2}(2)(10) + \frac{1}{2}(2)(-10) = 10 - 10 = 0$ m。
Displacement = area under $v$-$t$ curve. The line crosses zero at $t = 2$ s. Positive area ($0→2$) = 10 m, negative area ($2→4$) = −10 m. Net = 0 m.位移 = $v$-$t$ 曲线下面积。直线在 $t = 2$ s 处过零。正面积（$0\to 2$）= 10 m，负面积（$2\to 4$）= -10 m。合计 = 0 m。
3. A ball is launched horizontally at 15 m/s from a 20 m cliff. How far from the base does it land? ($g = 10$ m/s²)3. 小球以 15 m/s 水平方向从 20 m 高的悬崖发射。落地点距崖底多远？（$g = 10$ m/s²）
$30$ m
$20$ m
$45$ m
$15$ m
Correct! Fall time: $20 = \frac{1}{2}(10)t^2 \Rightarrow t = 2$ s. Range: $x = 15 \times 2 = 30$ m.正确！下落时间：$20 = \frac{1}{2}(10)t^2 \Rightarrow t = 2$ s。水平距离：$x = 15 \times 2 = 30$ m。
First find fall time from $h = \frac{1}{2}gt^2$: $t = 2$ s. Then $x = v_{x0} \times t = 15 \times 2 = 30$ m.先由 $h = \frac{1}{2}gt^2$ 求下落时间 $t = 2$ s，再算 $x = v_{x0} \times t = 15 \times 2 = 30$ m。
4. Which statement about inertial reference frames is correct?4. 关于惯性参考系（inertial reference frame），下列哪句正确？
Velocity is the same in all inertial frames所有惯性参考系中速度相同
Position is the same in all inertial frames所有惯性参考系中位置相同
Displacement is the same in all inertial frames所有惯性参考系中位移相同
Acceleration is the same in all inertial frames所有惯性参考系中加速度相同
Correct! This is Galilean invariance: acceleration is frame-independent across all inertial (non-accelerating) reference frames. Velocity and position are frame-dependent.正确！这就是伽利略不变性（Galilean invariance）：在所有非加速的惯性参考系中加速度都相同。速度和位置都与参考系有关。
Acceleration is invariant across inertial frames. Velocity depends on the observer's frame; only acceleration remains the same.加速度在惯性参考系之间不变。速度依赖于观察者所在参考系；只有加速度恒定。
5. If a car doubles its speed, its stopping distance (with constant braking force) becomes:5. 一辆车的速度翻倍（刹车力恒定），其刹车距离将变为：
Doubled2 倍
Quadrupled4 倍
Tripled3 倍
Unchanged不变
Correct! From $v^2 = v_0^2 + 2a\Delta x$, stopping distance $\Delta x = v_0^2/(2|a|)$. Since $\Delta x \propto v_0^2$, doubling $v_0$ quadruples $\Delta x$.正确！由 $v^2 = v_0^2 + 2a\Delta x$ 得刹车距离 $\Delta x = v_0^2/(2|a|)$。因为 $\Delta x \propto v_0^2$，$v_0$ 翻倍意味着 $\Delta x$ 变 4 倍。
Stopping distance is proportional to $v_0^2$. Double the speed → $2^2 = 4$ times the distance.刹车距离正比于 $v_0^2$。速度翻倍 → 距离变为 $2^2 = 4$ 倍。
Worked Example — Projectile Range and the 45° Result (FRQ Style)例题 —— 抛体射程与 45° 最优角（FRQ 风格）
A projectile is launched from ground level with initial speed $v_0$ at angle $\theta$ above the horizontal. Air resistance is negligible. (a) Derive the horizontal range $R(\theta)$ and the time of flight. (b) Show that $R$ is maximized at $\theta = 45^{\circ}$. (c) Compute $R$ for $v_0 = 20\;\text{m/s}$, $\theta = 30^{\circ}$, $g = 9.8\;\text{m/s}^2$.
Principle: 2-D projectile motion as independent constant-acceleration kinematics in $x$ (no force) and $y$ (gravity only); calculus optimization on the closed-form range.
$R$ depends on $\theta$ only through $\sin(2\theta)$, which has its maximum value of $1$ when $2\theta = 90^{\circ}$:$R$ 仅通过 $\sin(2\theta)$ 依赖 $\theta$，而 $\sin(2\theta)$ 在 $2\theta = 90^{\circ}$ 时达到最大值 $1$：
The second derivative is $-4v_0^2\sin(2\theta)/g < 0$ at $\theta = 45^{\circ}$, confirming a maximum:二阶导数 $-4v_0^2\sin(2\theta)/g$ 在 $\theta = 45^{\circ}$ 处小于零，确认是极大值：
$R_\text{max} = v_0^2/g = 400/9.8 \approx 40.8\;\text{m}$ for these inputs — the $\theta = 30^{\circ}$ result is correctly less. ✓在该数值条件下 $R_\text{max} = v_0^2/g = 400/9.8 \approx 40.8\;\text{m}$，而 $\theta = 30^{\circ}$ 算出的 35.3 m 确实小于该值。✓
Symmetry: $R(30^{\circ}) = R(60^{\circ})$ because $\sin 60^{\circ} = \sin 120^{\circ}$. Two angles always reach the same range (except $45^{\circ}$, the unique max). ✓对称性：$R(30^{\circ}) = R(60^{\circ})$，因为 $\sin 60^{\circ} = \sin 120^{\circ}$。两个互补角射程相同（$45^{\circ}$ 是唯一的最大值）。✓
Limit $\theta \to 0$ or $\theta \to 90^{\circ}$: $\sin(2\theta) \to 0$, so $R \to 0$ — consistent with launching nearly horizontally (no time aloft) or nearly vertically (no horizontal velocity). ✓极限 $\theta \to 0$ 或 $\theta \to 90^{\circ}$：$\sin(2\theta) \to 0$，故 $R \to 0$——和"几乎水平发射没有滞空时间"或"几乎竖直发射没有水平速度"的直觉一致。✓
Worked Example — Non-Constant Acceleration: Integrating $a(t)$ (FRQ Style)例题 —— 非恒定加速度：对 $a(t)$ 积分（FRQ 风格）
A particle moves along the $x$-axis with acceleration $a(t) = 6t - 12\;\text{m/s}^2$. At $t = 0$ the particle is at $x_0 = 5\;\text{m}$ with $v_0 = 0$. (a) Derive $v(t)$ and $x(t)$. (b) Find every time on $0 \le t \le 5\;\text{s}$ at which the particle is momentarily at rest. (c) Compute the total distance (not displacement) traveled on $0 \le t \le 5\;\text{s}$.
Principle: velocity and position from non-constant acceleration via direct integration; total distance requires partitioning the interval at sign changes of $v(t)$.
$$3t^2 - 12t = 3t(t - 4) = 0 \;\Rightarrow\; t = 0\;\text{s},\; t = 4\;\text{s}$$
Both solutions lie in $[0, 5]$, so the particle is at rest at $t = 0$ (the initial condition) and again at $t = 4\;\text{s}$.两根都落在 $[0, 5]$ 内，因此质点在 $t = 0$（初始条件）以及 $t = 4\;\text{s}$ 处瞬时静止。
(c) Total distance — partition at the sign change(c) 总路程 —— 在变号点处分段
$v(t) = 3t(t - 4)$ is negative on $(0, 4)$ and positive on $(4, 5)$, so the particle moves left then right. Sum the magnitudes of the two displacements:$v(t) = 3t(t - 4)$ 在 $(0, 4)$ 区间为负、$(4, 5)$ 区间为正，质点先向左后向右运动。把两段位移的绝对值相加：
Net displacement $x(5) - x(0) = -25\;\text{m}$, smaller in magnitude than the $39\;\text{m}$ total distance. The two only agree when $v$ does not change sign. ✓净位移 $x(5) - x(0) = -25\;\text{m}$，其大小小于 $39\;\text{m}$ 的总路程。只有当 $v$ 不变号时两者才相等。✓
A common slip is to compute $|x(5) - x(0)|$ and call it "distance." That gives the displacement magnitude, not the path length — the partition at $t = 4$ is the whole point. ✓常见错误：把 $|x(5) - x(0)|$ 直接当成"总路程"。那只是位移大小，并非路径长度——在 $t = 4$ 处分段正是为了避免该错误。✓