位置、速度与加速度

AP 物理1· AP 物理1 课程与考试说明 — 运动学· 14 分钟阅读

1. 核心运动学物理量★☆☆☆☆⏱ 3 min

本知识点是整个运动学的基础，运动学在AP物理1考试总分中占比10%-16%。本知识点的概念是从力学到圆周运动所有运动分析题的基础，因此准确识别物理量至关重要。

📘 定义

核心运动学物理量

核心物理量按属性分为标量（仅含大小）和矢量（同时含大小与方向）两类：

例:

Position (vector): Location relative to a defined origin
Displacement (vector): Net change in position, $Δ x = x_{f} - x_{i}$
Distance (scalar): Total path length traveled
Velocity (vector): Rate of change of position
Speed (scalar): Magnitude of velocity
Acceleration (vector): Rate of change of velocity

📐 例题

一名徒步者沿平直步道向东行走3公里，随后掉头向西朝起点方向行走1公里。取东为正方向，求徒步者的总路程和总位移。

1
总路程是所有路径段的标量和：
2
$3\ \text{km} + 1\ \text{km} = 4\ \text{km}$
3
位移是位置的矢量变化量，初始位置为 $x_{i} = 0$ ：
4
$x_f = 3 - 1 = 2\ \text{km}, \quad \Delta x = x_f - x_i = 2\ \text{km}$

2. 平均量与瞬时量★★☆☆☆⏱ 4 min

所有运动学物理量都可以分为平均量（在有限时间间隔内测量）和瞬时量（在单个时间点测量）两类。平均量的定义适用于所有运动，无论加速度是恒定还是变化的。

\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}, \quad \bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}

瞬时量是在单个时间点测得的物理量。从图像角度看，瞬时速度是 $x$ - $t$ 图像上切线的斜率，瞬时加速度是 $v$ - $t$ 图像上切线的斜率。在AP物理1考试中你可以直接从图像读取这些斜率，不需要用到微积分知识。

📐 例题

一名骑行者沿直线路径运动，位置函数为 $x (t) = 0.5 t^{2} + 2 t + 1$ ，其中 $x$ 单位为米， $t$ 单位为秒。求(a) $t = 1 s$ 到 $t = 5 s$ 之间的平均速度，以及(b) $t = 3 s$ 时的瞬时速度。

1
计算该时间区间的初始位置和末位置：
2
$x(1) = 0.5(1)^2 + 2(1) + 1 = 3.5\ \text{m}, \quad x(5) = 0.5(25) + 2(5) + 1 = 23.5\ \text{m}$
3
使用核心定义计算平均速度：
4
$\Delta x = 23.5 - 3.5 = 20\ \text{m}, \quad \Delta t = 5 - 1 = 4\ \text{s}, \quad \bar{v} = \frac{20}{4} = 5\ \text{m/s}$
5
对于瞬时速度，注意该位置函数符合匀加速运动的形式 $x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$ 。读取系数可得 $v_{0} = 2 m/s$ ，且 $\frac{1}{2} a = 0.5$ ，因此 $a = 1 m/s^{2}$ 。
6
代入匀加速运动公式 $v = v_{0} + a t$ ，取 $t = 3 s$ ：
7
$v = v_0 + a t = 2 + (1)(3) = 5\ \text{m/s}$

3. 物理量之间的图像关系★★☆☆☆⏱ 4 min

AP物理1非常侧重运动学物理量的图像解读考点。核心关系遵循两条简单规则：

Slope Rule: The slope of any kinematic graph equals the next quantity in the chain: $slope of x - t = v$ , $slope of v - t = a$
Area Rule: The net signed area under any kinematic graph equals the change in the previous quantity in the chain: $net area under v - t = Δ x$ , $net area under a - t = Δ v$

面积带有正负号：时间轴上方的面积为正，时间轴下方的面积为负，分别对应正的或负的速度/加速度。

📐 例题

沿平直轨道运动的玩具小车的 $v$ - $t$ 图像包含三段：(1) $t = 0$ 到 $t = 2 s$ ：位于 $v = 4 m/s$ 的水平线，(2) $t = 2 s$ 到 $t = 4 s$ ：从 $v = 4 m/s$ 到 $v = 0$ 的直线，(3) $t = 4 s$ 到 $t = 6 s$ ：从 $v = 0$ 到 $v = - 4 m/s$ 的直线。求小车从 $t = 0$ 到 $t = 6 s$ 的总位移。

1
计算每一段的带符号面积，因为 $v$ - $t$ 图像下的面积等于位移：
2
第一段（0到2秒）：矩形面积：
3
$2\ \text{s} \times 4\ \text{m/s} = 8\ \text{m}$
4
第二段（2到4秒）：三角形面积：
5
$0.5 \times 2\ \text{s} \times 4\ \text{m/s} = 4\ \text{m}$
6
第三段（4到6秒）：位于轴下方的三角形，因此面积为负：
7
$0.5 \times 2\ \text{s} \times (-4\ \text{m/s}) = -4\ \text{m}$
8
将所有面积相加得到总位移：
9
$\Delta x_{\text{total}} = 8 + 4 - 4 = 8\ \text{m}$

4. 匀加速运动学方程★★★☆☆⏱ 3 min

对于加速度恒定的运动（例如地球表面附近的自由落体、恒定刹车过程），我们可以推导出三个简化方程，大幅提升解题速度。这些方程仅在加速度恒定的条件下成立——如果加速度发生变化，请改用图像方法或核心定义求解。

\begin{align} v &= v_0 + a t \tag{1} \\ \Delta x &= v_0 t + \frac{1}{2} a t^2 \tag{2} \\ v^2 &= v_0^2 + 2 a \Delta x \tag{3} \end{align}

每个方程都省略了一个未知量，因此你总能选出与已知量匹配的方程，一步求解出目标未知量。

📐 例题

将一个小球从地面以初速度 $20 m/s$ 竖直向上抛出，取向上为正方向，重力加速度 $a = - 10 m/s^{2}$ ，求小球能到达的最大高度。

1
列出所有已知量和未知量：初速度 $v_{0} = 20 m/s$ ，末速度 $v = 0 m/s$ （小球在最高点处瞬时静止），加速度 $a = - 10 m/s^{2}$ ，待求未知量为位移 $Δ x$ （即最大高度）。
2
选择不含时间（无需额外求解的量）的运动学方程：即第3个运动学公式。
3
将已知数值代入该方程：
4
$0^2 = (20)^2 + 2(-10)\Delta x$
5
求解 $Δ x$ ：
6
$0 = 400 - 20\Delta x \rightarrow 20\Delta x = 400 \rightarrow \Delta x = 20\ \text{m}$

5. AP 风格概念检测★★★☆☆⏱ 4 min

✓ 快速检测

请完成以下AP风格习题，检验你对知识点的掌握程度：

The velocity of an object moving in a straight line is shown on a $v$ - $t$ graph as a straight line rising from $2 m/s$ at $t = 0$ to $10 m/s$ at $t = 4 s$ . What is the instantaneous acceleration of the object at $t = 2 s$ ?
- $2 m/s^{2}$
- $4 m/s^{2}$
- $5 m/s^{2}$
- $8 m/s^{2}$

📐 例题

一名驾驶员在郊区街道上以 $15 m/s$ 的速度行驶，此时一只猫突然冲到路面上。驾驶员的反应时间为 $0.6 s$ （即从看到猫到踩下刹车的时间间隔），踩下刹车后车辆获得大小为 $4.5 m/s^{2}$ 的恒定减速度。求从驾驶员看到猫到车辆完全停下的全过程中，汽车行驶的总距离。

1
将整个运动过程拆分为两个阶段：反应时间阶段（匀速运动，加速度为0）和刹车阶段（匀减速运动）。
2
反应时间阶段的行驶距离：
3
$d_1 = v_0 t_{\text{react}} = 15 \times 0.6 = 9\ \text{m}$
4
对于刹车阶段： $v_{0} = 15 m/s$ ， $v = 0$ （车辆已停下）， $a = - 4.5 m/s^{2}$ ：
5
使用不含时间的匀加速运动方程：
6
$0 = 15^2 + 2(-4.5)d_2 \rightarrow 0 = 225 - 9d_2 \rightarrow d_2 = 25\ \text{m}$
7
总停车距离：
8
$d_{\text{total}} = d_1 + d_2 = 9 + 25 = 34\ \text{m}$

6. 常见陷阱

错误做法:

Calculates average velocity as $(v_{i} + v_{f}) /2$ for motion with non-constant acceleration.

原因:

Students memorize this shortcut for constant acceleration and overgeneralize it to all motion.

正确做法:

Always calculate average velocity from the core definition $\overset{v}{ˉ} = Δ x /Δ t$ for any motion; only use the average-of-velocities shortcut when acceleration is explicitly constant.

错误做法:

Interprets the y-value of a velocity-time graph as position.

原因:

Students mix up graph axes, confusing position and velocity quantities.

正确做法:

Before interpreting any kinematic graph, label each axis, then remind yourself: 'slope of current = next quantity, area of current = previous quantity'.

错误做法:

Adds total path length to get displacement, giving a positive displacement when the object ends up left of its starting position.

原因:

Students confuse scalar distance with vector displacement.

正确做法:

Always calculate displacement as final position minus initial position, regardless of the path taken between them.

错误做法:

Only takes the positive square root when solving $v^{2} = v_{0}^{2} + 2 a Δ x$ , even when the object is moving in the negative direction.

原因:

Students assume velocity is always positive, forgetting velocity is a signed vector.

正确做法:

After taking the square root, always check the direction of motion to select the correct sign for your final answer.

错误做法:

Claims that zero acceleration means zero velocity.

原因:

Students confuse acceleration (rate of change of velocity) with velocity itself.

正确做法:

Always remember: zero acceleration means constant velocity (can be non-zero), zero velocity means instantaneous zero speed (can have non-zero acceleration, e.g., a ball at maximum height).

错误做法:

Treats all area under a $v$ - $t$ graph as positive when calculating net displacement.

原因:

Students think area is always positive, ignoring that negative velocity produces negative displacement.

正确做法:

Assign a negative sign to all area that lies below the time axis before summing for net change.

7. 速查表

Category	Formula	Notes
Displacement	$Δ x = x_{f} - x_{i}$	Vector net change; not equal to total distance
Average Velocity	$\overset{v}{ˉ} = Δ x /Δ t$	Valid for all motion, constant or not
Instantaneous Velocity	$v = slope of x - t$	Read off as the tangent slope on the graph; no calculus needed
Average Acceleration	$\overset{a}{ˉ} = Δ v /Δ t$	Valid for all motion
Instantaneous Acceleration	$a = slope of v - t$	Also equals the slope of the tangent to the $v$ - $t$ graph
Graphical Area Rule	$Δ x = net area under v - t; Δ v = net area under a - t$	Area below axis is negative, area above is positive
Constant Acceleration 1	$v = v_{0} + a t$	Valid only for constant $a$ ; omits $Δ x$
Constant Acceleration 2	$Δ x = v_{0} t + \frac{1}{2} a t^{2}$	Valid only for constant $a$ ; omits final velocity $v$
Constant Acceleration 3	$v^{2} = v_{0}^{2} + 2 a Δ x$	Valid only for constant $a$ ; omits time $t$

深入阅读

Syllabus官方AP物理1课程与考试说明官方运动学学习目标

下一步

Position, velocity, and acceleration are the foundational building blocks for all of kinematics, and for the entire AP Physics 1 course. Every topic that follows, from Newton’s laws of motion to energy, momentum, and circular motion, relies on your ability to correctly relate these three quantities to analyze motion. Without a solid understanding of how to interpret graphs of these quantities and apply the constant acceleration kinematic equations, you will struggle to set up and solve almost every free-response question on the exam, and many multiple-choice questions as well. Next, you will extend these 1-dimensional concepts to 2-dimensional motion, starting with projectile motion, where you separate horizontal and vertical motion into independent 1-dimensional kinematics problems that use all the rules you learned here.