恒定加速度运动学公式

AP 物理 1· AP 物理 1 CED — 运动学· 14 分钟阅读

1. 核心概念与标准符号★★☆☆☆⏱ 3 min

恒定加速度运动学公式是一组描述加速度大小和方向都不变的物体一维运动的代数关系，是AP物理1考试中最常见的基础运动场景。该主题占AP物理1考试总分的5-7%，是涉及力、能量和动量的更复杂问题的基础。

📘 定义

恒定加速度运动学

无需积分即可求解未知运动物理量的一组代数关系，仅当加速度大小和方向恒定时成立。

例:

近地自由落体（忽略空气阻力）、匀加速行驶的车辆

AP物理1教材中统一使用的标准符号如下：

$x_{0}$ = initial position, $x$ = final position
$v_{0}$ = initial velocity, $v$ = final velocity
$a$ = constant acceleration
$t$ = time elapsed (with initial time $t_{0} = 0$ for simplicity)
$Δ x = x - x_{0}$ = net displacement

2. 四个核心公式的推导★★☆☆☆⏱ 4 min

🔬 推导

目标:

从基本原理推导恒定加速度的全部四个核心运动学公式

起点:

恒定加速度和平均速度的定义

1
Start from the definition of constant acceleration, with $Δ t = t - 0 = t$ :
2
$a = \frac{v - v_0}{t}$
3
Rearranging gives the first equation, relating velocity to time:
4
$v = v_0 + at \tag{1}$
5
For constant acceleration, velocity changes linearly with time, so average velocity equals the average of initial and final velocity. By definition, average velocity also equals displacement over time:
6
$v_{\text{avg}} = \frac{v_0 + v}{2} = \frac{\Delta x}{t}$
7
Rearranging gives the second equation, relating displacement to average velocity and time:
8
$\Delta x = \frac{1}{2}(v_0 + v)t \tag{2}$
9
Substitute Equation (1) into Equation (2) to eliminate final velocity $v$ , and simplify to get the third equation, relating displacement to time:
10
$\Delta x = v_0 t + \frac{1}{2} a t^2 \tag{3}$
11
Finally, rearrange Equation (1) to solve for $t = \frac{v - v _{0}}{a}$ , substitute into Equation (2) to eliminate time, and simplify to get the fourth time-independent equation:
12
$v^2 = v_0^2 + 2 a \Delta x \tag{4}$

结果:

每个公式恰好省略五个运动学物理量( $v, v_{0}, a, Δ x, t$ )中的一个。解题时，始终选择包含所有已知量和仅一个未知量的公式。

📐 例题

一辆汽车从静止开始以 $2.0 m/s^{2}$ 的恒定加速度加速了5.0秒。该汽车在这段时间内的总位移是多少？

1
列出所有已知量和我们需要求解的未知量：
2
$v_{0} = 0 m/s$ (starting from rest), $a = 2.0 m/s^{2}$ , $t = 5.0 s$ , unknown $Δ x = ?$
3
Identify the correct equation: Equation (3) $Δ x = v_{0} t + \frac{1}{2} a t^{2}$ includes all three knowns and solves directly for $Δ x$ .
4
Substitute values into the equation:
5
$\Delta x = (0)(5.0) + \frac{1}{2}(2.0)(5.0)^2$
6
Calculate the final result:
7
$\Delta x = 25 \text{ m}$

Exam tip:

选择公式前一定要先列出你的已知量和未知量。这个简单步骤可以消除90%因选择了带有两个未知量的公式而导致的常见错误。

3. 矢量符号约定与自由落体★★★☆☆⏱ 3 min

所有运动学物理量（位移、速度、加速度）都是矢量，因此它们的符号完全取决于你在解题开始时选择的坐标系。一致性比选择哪个方向为正重要得多：只要每个矢量的符号都符合你的坐标系，你就会得到正确的结果。

For near-Earth free fall, acceleration due to gravity always points downward, regardless of the object's direction of motion: if upward is positive, $a = - g = - 9.8 m/s^{2}$ ; if downward is positive, $a = + g = + 9.8 m/s^{2}$ .

📐 例题

一个小球从地面竖直上抛，初速度为 $19.6 m/s$ 。取向上为正方向，求小球到达最大高度所需的时间。

1
Define the coordinate system: origin at ground level, upward = positive, so acceleration from gravity is $a = - 9.8 m/s^{2}$ .
2
List known quantities and unknown: $v_{0} = + 19.6 m/s$ , at maximum height the ball stops before falling back, so final velocity $v = 0 m/s$ , unknown $t = ?$ .
3
Choose the correct equation that includes all knowns and solves for $t$ : $v = v_{0} + a t$ .
4
Rearrange and solve for $t$ :
5
$t = \frac{v - v_0}{a} = \frac{0 - 19.6}{-9.8} = 2.0 \text{ s}$

Exam tip:

如果你得到负的时间或不合理的过大末速度，首先要检查的就是加速度相对于你所选坐标系的符号。

4. 图像解读★★★☆☆⏱ 3 min

AP物理1非常注重考察运动学公式和运动图像之间的联系。对于恒定加速度，每种图像都有可预测的形状，与核心公式直接对应：

Acceleration vs. time (a vs t): A horizontal line at $y = a$ , since acceleration is constant. The area under the a vs t graph equals $Δ v$ , matching $Δ v = a t$ from Equation (1).
Velocity vs. time (v vs t): A straight line, with slope equal to acceleration $a$ and y-intercept equal to $v_{0}$ , matching Equation (1). The area under the v vs t graph equals displacement $Δ x$ , matching Equation (2).
Position vs. time (x vs t): A parabola, since $x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$ from Equation (3). It opens upward for positive $a$ , downward for negative $a$ , and the slope of the tangent at any time equals instantaneous velocity.

📐 例题

The velocity vs time graph for a car slowing to a stop along a straight road is a straight line that goes from $(0, 10 m/s)$ to $(4 s, 2 m/s)$ . What is the total displacement of the car over the 4-second interval?

1
For v vs t graphs, displacement equals the area under the line, which can be confirmed with kinematic Equation (2).
2
List knowns: $v_{0} = 10 m/s$ , $v = 2 m/s$ , $t = 4 s$ , unknown $Δ x$ .
3
Substitute into Equation (2):
4
$\Delta x = \frac{1}{2}(10 + 2)(4) = 24 \text{ m}$
5
This matches the trapezoid area formula for graphs, confirming the result.

Exam tip:

For AP Physics 1 MCQ questions about motion graphs, you can almost always solve displacement problems faster using the area rule for v-t graphs than full algebraic substitution.

5. AP风格例题讲解★★★★☆⏱ 4 min

📐 例题

A tennis ball is dropped from rest from a height of 5.0 m above the ground. Air resistance is negligible. What is the approximate speed of the ball just before it hits the ground? Take downward as positive, $g = 9.8 m/s^{2}$ . Options: A) 5 m/s, B) 10 m/s, C) 15 m/s, D) 20 m/s

1
List knowns and unknown: $v_{0} = 0 m/s$ , $Δ x = 5.0 m$ , $a = 9.8 m/s^{2}$ , unknown final speed $v$ . Time is not given or required, so select the time-independent equation.
2
Substitute into $v^{2} = v_{0}^{2} + 2 a Δ x$ :
3
$v^2 = 0 + 2(9.8)(5.0) = 98$
4
Take the square root: $v = 98 \approx 10 m/s$ . The other options come from common errors: A from wrong time substitution, C/D from arithmetic errors. Correct answer is B.

📐 例题

A bicyclist is moving at a constant speed of $8.0 m/s$ when she passes a runner who is just starting to accelerate from rest at a constant $2.0 m/s^{2}$ in the same direction as the bicyclist. (a) How long does it take the runner to catch up to the bicyclist? (b) What is the speed of the runner when they meet? (c) Explain why the runner is moving faster than the bicyclist when they meet.

1
Part (a): Let $t$ = time after the bicyclist passes the runner. The bicyclist has zero acceleration, so their displacement is $Δ x_{b} = 8.0 t$ . The runner's displacement is $Δ x_{r} = \frac{1}{2} (2.0) t^{2} = t^{2}$ . When the runner catches up, displacements are equal:
2
$t^2 = 8.0 t \implies t(t - 8.0) = 0$
3
The non-zero solution is $t = 8.0 s$ .
4
Part (b): Use $v = v_{0} + a t$ for the runner:
5
$v = 0 + (2.0)(8.0) = 16 \text{ m/s}$
6
Part (c): For the first 4 seconds, the bicyclist is faster, so the distance between them increases. The runner's speed equals the bicyclist's at $t = 4 s$ , after which the runner is faster and the distance decreases until catch at $t = 8 s$ . By the time of catch, the runner has been faster for 4 seconds, so their speed is higher at meeting.

📐 例题

A police car is stopped at a traffic light. When the light turns green, a truck traveling at constant speed passes the police car going 20 m/s. The police car accelerates from rest at a constant $4 m/s^{2}$ to catch the truck. How far from the traffic light does the police car catch up to the truck?

1
Let $t$ = time after the light turns green. The truck's displacement is $Δ x_{T} = 20 t$ , and the police car's displacement is $Δ x_{P} = \frac{1}{2} (4) t^{2} = 2 t^{2}$ . When the police car catches the truck, displacements are equal:
2
$2 t^2 = 20 t \implies t = 10 \text{ s} (non-zero solution)$
3
Substitute back to find displacement from the traffic light:
4
$\Delta x_P = 2(10)^2 = 200 \text{ m}$

6. 常见陷阱

错误做法:

在非恒定加速度运动中使用 $v_{avg} = \frac{v _{0} + v}{2}$ 。

原因:

这个关系仅在加速度恒定时成立，因为只有加速度不变时速度才随时间线性变化。

正确做法:

使用任何运动学公式前，一定要确认题目说明加速度是恒定的；对于非恒定加速度，仅使用 $v_{avg} = \frac{总位移}{总时间}$ 。

错误做法:

在自由落体问题中选择向下为正方向时，加速度仍使用 $- g$ 。

原因:

学生死记'重力加速度是负的'，没有将符号和自己选择的坐标系联系起来。

正确做法:

选择坐标系后，代入任何公式前，都要根据方向明确写出加速度的符号。

错误做法:

当物体反向运动时，将净位移当成总路程上报。

原因:

学生混淆净位移和总路程；运动学公式仅给出净位移，不是各个方向运动路程的和。

正确做法:

如果题目要求总路程，将运动拆分为方向反转前后的分段（反转处v=0），计算每段位移的大小，然后相加。

错误做法:

对于从地面上抛后落回地面的小球，将到达最大高度的时间当成总飞行时间。

原因:

学生解出到达最大高度的时间后就停止了，忘记小球还要落回出发高度。

正确做法:

如果发射和落地高度相等，要求总飞行时间，可以将到达最大高度的时间翻倍，或者直接利用位移公式中 $Δ x = 0$ 求解。

错误做法:

错误整理 $v^{2} = v_{0}^{2} + 2 a Δ x$ ，导致根号下出现负值。

原因:

学生代数整理时太匆忙，将 $v_{0}^{2}$ 移到了等式错误的一侧。

正确做法:

始终逐步整理：先得到 $v^{2} - v_{0}^{2} = 2 a Δ x$ ，即 $v^{2} = v_{0}^{2} + 2 a Δ x$ ，再开平方。

7. 速查表

公式/关系	省略物理量	要点
$v = v_{0} + a t$	$Δ x$	仅适用于恒定加速度
$Δ x = \frac{1}{2} (v_{0} + v) t$	$a$	仅适用于恒定加速度
$Δ x = v_{0} t + \frac{1}{2} a t^{2}$	$v$	仅适用于恒定加速度
$v^{2} = v_{0}^{2} + 2 a Δ x$	$t$	仅适用于恒定加速度
$a = \pm 9.8 m/s^{2}$	N/A	自由落体：符号取决于坐标系，向上为正时取负
a vs t graph (constant $a$ )	N/A	水平线；图下面积 = 速度变化量 $Δ v$
v vs t graph (constant $a$ )	N/A	斜率为 $a$ 的直线；图下面积 = 位移 $Δ x$
x vs t graph (constant $a$ )	N/A	抛物线； $t$ 时刻切线斜率 = 瞬时速度

真题中的出现

AI 根据考纲规律估算的考点位置,请对照官方真题核实准确性。仅作复习重点参考。

2023 · MCQ
自由落体速度计算
2022 · FRQ
追及运动问题

下一步

恒定加速度运动学公式是AP物理1几乎所有运动主题的基础，首先就是抛体运动：在抛体运动中我们将运动分解为独立的水平和竖直分量，每个分量都具有恒定加速度（水平分量加速度为零，竖直分量加速度为g）。如果不掌握本主题的符号约定和公式选择，你就无法正确求解抛体运动问题，而抛体运动占AP考试运动学单元的很大一部分。本主题也直接为动力学（牛顿定律）打下基础：在动力学中你会用牛顿第二定律计算加速度，然后用这里的运动学公式求出物体的最终运动。这种核心联系意味着恒定加速度运动学几乎用于AP物理1的所有单元，从动量到圆周运动都有涉及。