AP Physics 1 · Concept 1.3AP 物理 1 · 概念1.3
Uniformly Accelerated Motion匀变速直线运动
Δx = v₀t + ½at²
An object is in uniformly accelerated motion when its velocity changes by the same amount every second. Its acceleration is constant.当物体的速度每秒变化量相同时，该物体处于匀变速运动状态。此时，其加速度是恒定的。
a = Δv / Δt = constant常量
With constant acceleration, three equations describe the whole motion:在加速度恒定时，有三个方程描述整个运动过程：
- v = v₀ + atvelocity after time t时间 t 后的速度
- Δx = v₀t + ½at²displacement after time t时间 t 后的位移
- v² = v₀² + 2aΔxwhen time is not given当时间未给定时
Displacement is the area under the velocity–time graph. With constant acceleration the graph is a straight line, so the area splits into two simple shapes.位移是速度-时间图下的面积。在加速度恒定时，该图为直线，因此该面积可分解为两个简单的图形。
The blue rectangle is v₀t. The gold triangle is ½at². Their sum is the equation.蓝色矩形为 v₀t。金色三角形为 ½at²。两者的和即为该方程。
A cart moving at 4.0 m/s speeds up at 2.0 m/s² for 3.0 s. Find its final velocity and its displacement.一辆以 4.0 m/s 速度运动的小车，以 2.0 m/s² 的加速度加速 3.0 s。求其末速和位移。
- Known:已知： v₀ = 4.0 m/s, a = 2.0 m/s², t = 3.0 s
- Velocity:末速度： v = 4.0 + (2.0)(3.0) = 10 m/s
- Displacement:位移： Δx = (4.0)(3.0) + ½(2.0)(3.0)² = 21 m
Check验证Average velocity (4 + 10) / 2 = 7 m/s, and 7 × 3 = 21 m. Same answer.平均速度 (4 + 10) / 2 = 7 m/s，且 7 × 3 = 21 m。结果一致。
- !Using these equations when acceleration changes当加速度发生变化时使用这些公式They only hold for constant acceleration. Otherwise use graphs or calculus.这些公式仅在加速度恒定时成立。否则应使用图像或微积分。
- !Skipping the sign convention省略符号约定Pick a positive direction first. A ball thrown upward has a = −9.8 m/s² the whole time.先选定一个正方向。向上抛出的球在整个过程中加速度 a = −9.8 m/s²。
- !Thinking a = 0 at the top误以为顶点处 a = 0At the top v = 0, but gravity still acts, so a is still −9.8 m/s².在顶点处 v = 0，但重力仍在起作用，因此 a 仍为 −9.8 m/s²。
