📚 Using Differentiation to Solve Velocity and Acceleration | 利用微分法求解速度与加速度
In many examination problems, a particle moves along a straight line and its position is given as a function of time. Differentiation provides a direct and powerful method for finding the velocity and acceleration from this position function. If s(t) is the displacement at time t, then the velocity is the first derivative and the acceleration is the second derivative.
在许多考试问题中，质点沿直线运动，其位置是时间的函数。微分法为我们提供了一种直接而有力的方法，可以从位置函数求出速度和加速度。若 s(t) 表示 t 时刻的位移，那么速度是位移的一阶导数，加速度是二阶导数。
This article will guide you through the key definitions, worked examples, common pitfalls, and exam-style practice questions. By the end, you should be able to move confidently between position, velocity and acceleration using differentiation.
本文将带你梳理核心定义、典型例题、常见易错点以及考试风格的练习题。学完之后，你应该能够熟练地利用微分法在位移、速度和加速度之间进行转换。
1. Displacement, Velocity and Acceleration: Definitions | 位移、速度与加速度的定义
Displacement s(t) describes the position of a particle relative to a fixed origin. It is a vector quantity, so it can be positive, negative or zero depending on the direction from the origin.
位移 s(t) 描述质点相对于固定原点的位置。它是矢量，因此根据相对原点的方向，位移可以为正、为负或为零。
Velocity is the rate of change of displacement with respect to time. The instantaneous velocity is the limit of the average velocity as the time interval tends to zero:
速度是位移对时间的变化率。瞬时速度是时间间隔趋于零时平均速度的极限：
v(t) = ds/dt = s'(t)
Acceleration is the rate of change of velocity with respect to time:
加速度是速度对时间的变化率：
a(t) = dv/dt = v'(t) = d²s/dt² = s”(t)
The SI units are metres per second (m/s) for velocity and metres per second squared (m/s²) for acceleration.
速度的国际单位是米每秒（m/s），加速度的国际单位是米每二次方秒（m/s²）。
2. Displacement versus Distance | 位移与路程的区别
Displacement is the change in position, so it only depends on the starting point and the finishing point. Distance is the total length of the path actually travelled, so it is always positive or zero.
位移是位置的变化量，只取决于起点和终点。路程是实际经过的总长度，因此始终大于或等于零。
For example, if a particle moves from s = 0 to s = 10 and then back to s = 4, its final displacement is 4 m, but the total distance travelled is 10 + 6 = 16 m.
例如，若质点从 s = 0 运动到 s = 10，再回到 s = 4，则其最终位移为 4 m，但总路程为 10 + 6 = 16 m。
When working with a position function, the displacement between t = t₁ and t = t₂ is s(t₂) − s(t₁). To find the total distance, you must split the motion at every instant when the velocity changes sign, because that is when the particle reverses direction.
使用位置函数时，从 t = t₁ 到 t = t₂ 的位移为 s(t₂) − s(t₁)。要求总路程，必须在速度改变符号的每一个时刻将运动分段，因为此时质点改变了运动方向。
3. The Differentiation Chain: s → v → a | 微分链：由位移到速度再到加速度
The relationship between displacement, velocity and acceleration is a beautiful chain of derivatives. Once you know s(t), you can differentiate to find v(t), and then differentiate again to find a(t).
位移、速度和加速度之间的关系是一条优美的导数链。一旦知道 s(t)，你可以通过求导得到 v(t)，再求导得到 a(t)。
s(t) → v(t) = ds/dt → a(t) = dv/dt
This means that the acceleration is the second derivative of displacement with respect to time. It also means that the gradient of a velocity-time graph gives acceleration, and the gradient of a displacement-time graph gives velocity.
这意味着加速度是位移对时间的二阶导数。同时，速度-时间图像的斜率给出加速度，位移-时间图像的斜率给出速度。
If the position is given as a polynomial, you apply the power rule term by term. For example, if s = t⁴, then v = 4t³ and a = 12t².
如果位置函数是多项式，你可以逐项使用幂函数求导法则。例如，若 s = t⁴，则 v = 4t³，a = 12t²。
4. Worked Example 1: One-Dimensional Motion | 例题一：一维直线运动
A particle moves along a straight line with displacement given by:
一个质点沿直线运动，其位移为：
s(t) = t³ − 6t² + 9t + 2
where s is in metres and t is in seconds. Find the velocity and acceleration functions.
其中 s 以米为单位，t 以秒为单位。求速度函数和加速度函数。
Differentiate to find the velocity:
求导得到速度：
v(t) = ds/dt = 3t² − 12t + 9 = 3(t − 1)(t − 3)
Differentiate again to find the acceleration:
再次求导得到加速度：
a(t) = dv/dt = 6t − 12 = 6(t − 2)
At t = 0, we have s(0) = 2 m, v(0) = 9 m/s and a(0) = −12 m/s². This means the particle starts at position 2 m, moving in the positive direction but slowing down.
在 t = 0 时，s(0) = 2 m，v(0) = 9 m/s，a(0) = −12 m/s²。这意味着质点从位置 2 m 处开始，沿正方向运动但正在减速。
5. Turning Points and Changes of Direction | 转向点与方向改变
A particle is momentarily at rest when v(t) = 0. These instants are called turning points if the velocity changes sign.
当 v(t) = 0 时，质点瞬时静止。若速度在该时刻改变符号，则这些时刻称为转向点。
In the previous example, v(t) = 3(t − 1)(t − 3). Therefore v = 0 when t = 1 or t = 3.
在前一例题中，v(t) = 3(t − 1)(t − 3)。因此当 t = 1 或 t = 3 时，v = 0。
For t < 1, both factors are negative, so v > 0. For 1 < t < 3, the factor (t − 1) is positive and (t − 3) is negative, so v < 0. For t > 3, v > 0.
当 t < 1 时，两个因子均为负，因此 v > 0。当 1 < t < 3 时，(t − 1) 为正而 (t − 3) 为负，因此 v < 0。当 t > 3 时，v > 0。
Thus the particle reverses direction at t = 1 and again at t = 3. The corresponding displacements are:
因此，质点在 t = 1 和 t = 3 时改变方向。对应的位移为：
s(1) = 1 − 6 + 9 + 2 = 6 m
s(3) = 27 − 54 + 27 + 2 = 2 m
At t = 1, the displacement is a local maximum; at t = 3, it is a local minimum. This matches the sign pattern of the velocity.
在 t = 1 时，位移达到局部极大值；在 t = 3 时，位移达到局部极小值。这与速度的符号变化模式一致。
We can also calculate the total distance travelled between t = 0 and t = 4. The particle moves forward from s = 2 to s = 6, then backward from s = 6 to s = 2, then forward from s = 2 to s(4) = 6. The total distance is 4 + 4 + 4 = 12 m.
我们还可以计算从 t = 0 到 t = 4 的总路程。质点先从 s = 2 前进到 s = 6，再从 s = 6 后退到 s = 2，然后从 s = 2 前进到 s(4) = 6。总路程为 4 + 4 + 4 = 12 m。
6. Finding Maximum or Minimum Velocity | 求速度的最大值或最小值
To find stationary values of velocity, you set the acceleration equal to zero. This is because acceleration is the derivative of velocity.
要求速度的极值，需要令加速度为零。这是因为加速度是速度的导数。
Using the same example, a(t) = 6t − 12. Setting a = 0 gives t = 2. At this instant:
使用同一个例题，a(t) = 6t − 12。令 a = 0，得到 t = 2。此时：
v(2) = 3(2)² − 12(2) + 9 = 12 − 24 + 9 = −3 m/s
To determine whether this is a maximum or minimum, look at the second derivative of velocity. Since v”(t) = a'(t) = 6 > 0, the velocity has a minimum at t = 2. The speed at this instant is 3 m/s.
要判断这是最大值还是最小值，需要考察速度的二阶导数。由于 v”(t) = a'(t) = 6 > 0，速度在 t = 2 处取得极小值。此时速率为 3 m/s。
Remember that a minimum velocity can still be negative. Speed is the magnitude of velocity, so it is always non-negative.
请记住，最小速度仍可能为负值。速率是速度的大小，因此始终非负。
7. Vertical Motion Under Gravity | 重力作用下的竖直运动
When a particle is projected vertically upwards, the acceleration due to gravity is approximately constant. If we take upward as the positive direction, then:
当质点竖直向上抛出时，重力加速度近似恒定。若取向上为正方向，则：
a = −g
where g ≈ 9.8 m/s². Starting from ground level with initial velocity u, differentiation gives the following results:
其中 g ≈ 9.8 m/s²。从地面以初速度 u 竖直上抛时，由微分可得：
v(t) = u − gt
s(t) = ut − ½gt²
At the highest point, the particle is momentarily at rest, so v = 0. Therefore the time to reach maximum height is:
在最高点，质点瞬时静止，因此 v = 0。所以到达最大高度的时间为：
t = u/g
Substituting this into the displacement equation gives the maximum height:
将其代入位移方程，得到最大高度：
s_max = u²/(2g)
For example, if u = 20 m/s, then t = 20/9.8 ≈ 2.04 s and s_max = 400/(19.6) ≈ 20.4 m.
例如，若 u = 20 m/s，则 t = 20/9.8 ≈ 2.04 s，s_max = 400/19.6 ≈ 20.4 m。
8. Motion in Two Dimensions | 二维运动
Differentiation also works for projectile or planar motion. If the position vector is:
微分法同样适用于抛体运动或平面运动。若位置矢量为：
r(t) = x(t)i + y(t)j
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