📚 Variable Acceleration: Mastering Calculus-Based Kinematics | 变加速运动：掌握微积分运动学
In A-Level Mechanics, the equations of motion for constant acceleration (SUVAT) provide a foundation, but real-world motion often involves acceleration that changes with time, velocity, or displacement. This topic, variable acceleration, requires the use of calculus to connect acceleration, velocity, and displacement. Mastering these techniques is essential for success in Edexcel Mechanics.
在A-Level力学中，匀加速运动方程(SUVAT)提供了基础，但现实中的运动往往涉及随时间、速度或位移变化的加速度。这个变加速度专题需要使用微积分将加速度、速度和位移联系起来。掌握这些技巧对在Edexcel力学中取得成功至关重要。
1. What Is Variable Acceleration? | 什么是变加速度？
or displacement s. For example, a = 3t + 2 (function of t), a = 4 - v (function of v), or a = 6s (function of s).
在匀加速问题中，加速度a是一个固定值。在变加速度中，a表示为时间t、速度v或位移s的函数。例如，a = 3t + 2 (t的函数)，a = 4 - v (v的函数)，或 a = 6s (s的函数)。
quivalently, using integration, velocity is the integral of acceleration, and displacement is the integral of velocity.
解决这些问题的关键在于理解导数关系：v = ds/dt 和 a = dv/dt。等价地，使用积分，速度是加速度的积分，位移是速度的积分。
Since these relationships involve derivatives and integrals, you will often need to find constants of integration using given initial conditions (e.g., initial velocity, initial position).
由于这些关系涉及导数和积分，你经常需要使用给定的初始条件（如初速度、初始位置）来求出积分常数。
2. From Acceleration to Velocity by Integration | 从加速度到速度的积分
dt. The constant of integration is found using the initial velocity, typically when t = 0.
如果加速度是时间的函数 a = f(t)，则在时间t的速度 v(t) 由 v(t) = ∫ a(t) dt 获得。积分常数通过初速度求出，通常当t=0时。
v(t) = ∫(6t+2) dt = 3t² + 2t + C. Substituting v(0)=3 gives C=3, so v = 3t² + 2t + 3.
例如，如果 a = 6t + 2，且在t=0时v=3，则 v(t) = ∫(6t+2) dt = 3t² + 2t + C。代入v(0)=3得C=3，因此 v = 3t² + 2t + 3。
v(t) = ∫ a dt = ∫ f(t) dt
v(t) = ∫ a dt = ∫ f(t) dt
3. From Velocity to Displacement | 从速度到位移
∫ v dt. Again, use an initial condition, such as s(0)=0 or a given starting position.
一旦速度已知为时间的函数 v = g(t)，位移s可以通过积分找到：s = ∫ v dt。同样，使用初始条件，如s(0)=0或给定的起始位置。
int; (3t² + 2t + 3) dt = t³ + t² + 3t + D. Because s=0 when t=0, D=0, so s = t³ + t² + 3t.
继续上例，v = 3t² + 2t + 3，且 s(0) = 0。则 s(t) = ∫ (3t² + 2t + 3) dt = t³ + t² + 3t + D。因为当t=0时s=0，得D=0，所以 s = t³ + t² + 3t。
Always remember to add the constant of integration and use the initial condition to determine its value. This applies even when dealing with vectors.
始终记得加上积分常数，并利用初始条件确定其值。即使处理矢量时也如此。
4. Using Initial Conditions Effectively | 有效使用初始条件
Common initial conditions include "at t=0, s=0", "from rest" (v=0 at t=0), or "at the origin" to find the constant of integration. For vector forms, apply to each component separately.
常见的初始条件包括 "t=0时，s=0"，"从静止开始" (t=0时v=0)，或 "在原点"。始终代入t=0和给定值以求出积分常数。对于矢量形式，分别应用于每个分量。
If a particle is "released from rest" 0. This gives two constants to be determined from the velocity and displacement functions.
如果质点从原点 "由静止释放"，则在t=0时v=0且s=0。这有两个常数需要由速度和位移函数确定。
Be careful with phrasing: "initially at the origin" means s=0 at t=0; "moving with speed 5 m/s" means the speed is 5, but the direction (sign of velocity) must be deduced from context.
注意措辞："初始在原点" 意味着t=0时s=0；"以5 m/s的速度运动" 意味着速率为5，但方向（速度的正负）需从上下文推断。
5. Acceleration as a Function of Velocity: a = f(v) | 加速度作为速度的函数：a = f(v)
ive form a = v (dv/ds), because we can separate variables. For example, if a = k v, then dv/dt = k v, which can be solved to give exponential velocity.
当加速度以速度的函数给出时，a = f(v)，使用 a = dv/dt，但这会导致微分方程。我们常使用替代形式 a = v (dv/ds)，因为可以分离变量。例如，若 a = k v，则 dv/dt = k v，可解得指数形式的速度。
then separate variables: (v / f(v)) dv = ds. Integrate both sides to find s in terms of v.
要将 a = f(v) 转换为v与s的关系，使用 a = v dv/ds。整理得 v dv = a ds = f(v) ds，然后分离变量：(v / f(v)) dv = ds。两边积分求得s关于v的表达式。
eaching a certain speed. In such cases, integrating a = v dv/ds is the most direct method.
考题可能要求达到某速度前经过的距离。这种情况下，对 a = v dv/ds 积分是最直接的方法。
6. Acceleration as a Function of Displacement: a = f(s) | 加速度作为位移的函数：a = f(s)
v = a ds. Integrate both sides: ∫ v dv = ∫ a ds. This yields a relationship between v and s.
如果 a = f(s)，我们不能直接对t积分，因为s不是变量。改用链式法则：a = dv/dt = v dv/ds。因此，v dv = a ds。两边积分：∫ v dv = ∫ a ds。这给出了v和s之间的关系。
½ v² = ∫ f(s) ds + C
½ v² = ∫ f(s) ds + C
C. Using v=0 when s=0 gives C=0, so v² = 6s², hence v = √6 s (taking positive root).
例如，如果 a = 6s 且质点从s=0处静止出发，则 v dv = 6s ds。积分得：½ v² = 3s² + C。利用v=0时s=0得C=0，所以 v² = 6s²，因此 v = √6 s（取正根）。
7. Finding Maximum Velocity and Turning Points | 求最大速度和转折点
Velocity is at a maximum or minimum when acceleration a = dv/dt = 0. To determine if it' a change in direction occurs when v = 0. This is critical for total distance calculations.
当加速度 a = dv/dt = 0 时，速度达到最大或最小值。要判断是否为最大值，检查a在两侧的符号或使用二阶导数（急动度）。另外，当v=0时方向改变，这对总路程计算至关重要。
ng a=0 gives t=2. Since a changes from positive to negative at t=2, v is maximum at t=2, vmax = 12(2) - 3(4) + 5 = 17 m/s.
例如，若 v = 12t - 3t² + 5，则 a = 12 - 6t。令a=0得t=2。由于a在t=2处由正变负，v在t=2处最大，vmax = 12(2) - 3(4) + 5 = 17 m/s。
article changes direction, so the displacement integral must be split at these times for total distance.
转折点（方向反转）通过求解v=0找到。质点改变方向，因此总路程的积分需在这些时刻分段。
8. Displacement vs Total Distance Travelled | 位移与总路程
absolute values of displacement over intervals. You must find times when v=0, split the integral, and integrate the absolute velocity or integrate velocity and take absolute values of each segment's area.
位移是速度对时间的积分；若运动方向为负，位移可为负。总路程是各区间位移绝对值之和。你必须找到v=0的时刻，分段积分，对速度绝对值积分或对每段速度积分后取绝对值。
en v=0 at t=1 and t=3. For a time interval from t=0 to t=4, split the integral: distance = |∫₀¹ v dt| + |∫₁³ v dt| + |∫₃⁴ v dt|. This yields total distance, not net displacement.
若 v(t) = t² - 4t + 3，则t=1和t=3时v=0。对于从t=0到t=4的时间区间，将积分分段：路程 = |∫₀¹ v dt| + |∫₁³ v dt| + |∫₃⁴ v dt|。这样得到总路程，而非净位移。
t)=0 within the given interval. Missing a turning point can lead to an incorrect distance.
始终检查在给定区间内v(t)=0的多个根。遗漏转折点会导致错误的路程。
9. Vector Variable Acceleration (2D) | 二维矢量变加速度
e acceleration means components are functions of t. Integrate each component separately to find velocity and position vectors.
在二维中，位置、速度和加速度是向量，例如 r = xi + yj，v = (dx/dt)i + (dy/dt)j，a = (d²x/dt²)i + (d²y/dt²)j。变加速度意味着分量是t的函数。分别对每个分量积分以求得速度向量和位置向量。
₁)i + (2t + C₂)j. Using initial conditions gives C₁=1, C₂=-1, so v = (1.5t²+1)i + (2t-1)j.
如果 a = (3t i + 2 j) m/s²，且在t=0时 v = (i - j) m/s，r = 0，则积分得：v = (∫3t dt)i + (∫2 dt)j = (1.5t² + C₁)i + (2t + C₂)j。使用初始条件得C₁=1，C₂=-1，所以 v = (1.5t²+1)i + (2t-1)j。
position vector r. Distance travelled in 2D is the integral of the magnitude of velocity, |v| = √(vx² + vy²), over time.
再次积分得到位置向量 r。二维中经过的路程是速度大小 |v| = √(vx² + vy²) 对时间的积分。
10. Exam-Style Problem and Common Mistakes | 考试题型与常见错误
A typical question: " city is 3 m/s and it is at the origin. Find the distance travelled in the first 4 seconds." Solution involves finding v(t), setting v=0 to find turning points, then integrating |v|.
典型问题："一个质点沿直线运动，加速度 a = (12 - 6t) m/s²。初始速度为3 m/s，在原点。求前4秒内经过的路程。" 解法包括求v(t)，令v=0求转向点，然后积分|v|。
; √5. Only t = 2 + √5 ≈ 4.236 is beyond 4 seconds, so within [0,4] the velocity does not change sign? Check: at t=0, v=3>0; at t=4, v=12(
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