恒定加速度运动学公式 — AP 物理 1
AP 物理 1 · AP 物理 1 CED 第一单元:运动学 · 14 min read
从基本原理推导恒定加速度的四个核心运动学公式
针对一维运动问题选择正确的运动学公式
对所选坐标系应用一致的矢量符号约定
解读恒定加速度运动的运动图像
解决AP物理1常见问题,包括自由落体和追及运动
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。
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。
Example: 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)
$\Delta x = x - x_0$ = net displacement
2. 四个核心公式的推导 ★★☆☆☆ ⏱ 4 min
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, $a = -g = -9.8 \text{ m/s}^2$; if downward is positive, $a = +g = +9.8 \text{ m/s}^2$.
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 $\Delta v$, matching $\Delta v = at$ 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 $\Delta x$, 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.
The velocity vs time graph for a car slowing to a stop along a straight road is a straight line that goes from $(0, 10 \text{ m/s})$ to $(4 \text{ s}, 2 \text{ m/s})$. What is the total displacement of the car over the 4-second interval?
For v vs t graphs, displacement equals the area under the line, which can be confirmed with kinematic Equation (2).
List knowns: $v_0 = 10 \text{ m/s}$, $v = 2 \text{ m/s}$, $t = 4 \text{ s}$, unknown $\Delta x$.
Substitute into Equation (2):
\Delta x = \frac{1}{2}(10 + 2)(4) = 24 \text{ m}
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
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, $g = 9.8 \text{ m/s}^2$. Options: A) 5 m/s, B) 10 m/s, C) 15 m/s, D) 20 m/s
List knowns and unknown: $v_0 = 0 \text{ m/s}$, $\Delta x = 5.0 \text{ m}$, $a = 9.8 \text{ m/s}^2$, unknown final speed $v$. Time is not given or required, so select the time-independent equation.
Substitute into $v^2 = v_0^2 + 2a\Delta x$:
v^2 = 0 + 2(9.8)(5.0) = 98
Take the square root: $v = \sqrt{98} \approx 10 \text{ m/s}$. The other options come from common errors: A from wrong time substitution, C/D from arithmetic errors. Correct answer is B.
A bicyclist is moving at a constant speed of $8.0 \text{ m/s}$ when she passes a runner who is just starting to accelerate from rest at a constant $2.0 \text{ m/s}^2$ 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.
Part (a): Let $t$ = time after the bicyclist passes the runner. The bicyclist has zero acceleration, so their displacement is $\Delta x_b = 8.0 t$. The runner's displacement is $\Delta x_r = \frac{1}{2}(2.0)t^2 = t^2$. When the runner catches up, displacements are equal:
t^2 = 8.0 t \implies t(t - 8.0) = 0
The non-zero solution is $t = 8.0 \text{ s}$.
Part (b): Use $v = v_0 + at$ for the runner:
v = 0 + (2.0)(8.0) = 16 \text{ m/s}
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 $t=4 \text{ s}$, after which the runner is faster and the distance decreases until catch at $t=8 \text{ s}$. By the time of catch, the runner has been faster for 4 seconds, so their speed is higher at meeting.
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 $4 \text{ m/s}^2$ to catch the truck. How far from the traffic light does the police car catch up to the truck?
Let $t$ = time after the light turns green. The truck's displacement is $\Delta x_T = 20 t$, and the police car's displacement is $\Delta x_P = \frac{1}{2}(4)t^2 = 2 t^2$. When the police car catches the truck, displacements are equal:
2 t^2 = 20 t \implies t = 10 \text{ s} (non-zero solution)
Substitute back to find displacement from the traffic light:
\Delta x_P = 2(10)^2 = 200 \text{ m}
Common Pitfalls
Why: 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。
Why: 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。
Why: 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。
Why: 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。
Why: 学生代数整理时太匆忙,将$v_0^2$移到了等式错误的一侧。
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