One-Dimensional Motion with Variable Acceleration | 一维变加速度运动分析

📚 One-Dimensional Motion with Variable Acceleration | 一维变加速度运动分析

Variable acceleration is the natural next step after constant acceleration. In one dimension, acceleration may change with time, velocity, or displacement, and calculus replaces the familiar SUVAT equations.

变加速度是学习完匀加速直线运动后的自然延伸。在一维问题中，加速度可以随时间、速度或位移而改变，微积分将取代我们熟悉的匀变速公式。

1. Why Variable Acceleration Matters | 为何研究变加速度

Constant acceleration is an idealisation. Real objects experience air resistance, springs, and non-uniform fields, so their acceleration is often a function of time \(t\), velocity \(v\), or displacement \(x\).

常加速度是理想模型。真实物体受到空气阻力、弹簧力或非均匀场作用时，加速度往往是时间 \(t\)、速度 \(v\) 或位移 \(x\) 的函数。

If acceleration depends on \(t\), we integrate twice to find velocity and position.

如果加速度依赖于 \(t\)，我们通过两次积分得到速度和位置。

If acceleration depends on \(v\), we usually separate variables and integrate.

如果加速度依赖于 \(v\)，我们通常采用分离变量法再积分。

If acceleration depends on \(x\), the identity \(a = v\frac{dv}{dx}\) is the most powerful tool.

如果加速度依赖于 \(x\)，恒等式 \(a = v\frac{dv}{dx}\) 是最有力的工具。

2. The Fundamental Equations of Variable Acceleration | 变加速度的基本关系

By definition, acceleration is the rate of change of velocity, and velocity is the rate of change of position:

根据定义，加速度是速度的变化率，速度是位置的变化率：

a = dv/dt = d²x/dt², v = dx/dt

When \(a\) is known as a function of time, the velocity is found by the first integral:

当 \(a\) 已知为时间的函数时，速度由第一次积分求得：

v(t) = u + ∫ a(t) dt

Here \(u\) is the velocity at \(t = 0\). The position is then the second integral:

其中 \(u\) 是 \(t = 0\) 时的速度。位置再由第二次积分得到：

x(t) = x₀ + ∫ v(t) dt

Every integration introduces a constant, which must be fixed using the initial conditions.

每一次积分都会产生积分常数，必须利用初始条件确定这些常数。

3. Acceleration as a Function of Time | 加速度为时间的函数

The simplest variable acceleration is a linear function of time. Suppose \(a = kt\), where \(k\) is a constant with units m/s³.

最简单的变加速度是时间的线性函数。设 \(a = kt\)，其中 \(k\) 是单位为 m/s³ 的常数。

Integrating once gives the velocity:

积分一次得到速度：

v(t) = u + ∫ kt dt = u + ½kt²

Integrating again gives the position:

再积分一次得到位置：

x(t) = x₀ + ut + (1/6)kt³

For example, if \(k = 2\) m/s³, \(u = 0\) and \(x₀ = 0\), then at \(t = 3\) s we have \(v = ½ × 2 × 3² = 9\) m/s and \(x = (1/6) × 2 × 3³ = 9\) m.

例如，若 \(k = 2\) m/s³，\(u = 0\)，\(x₀ = 0\)，则在 \(t = 3\) s 时，\(v = ½ × 2 × 3² = 9\) m/s，\(x = (1/6) × 2 × 3³ = 9\) m。

4. Acceleration as a Function of Velocity | 加速度为速度的函数

When air resistance is involved, acceleration often depends on velocity. The simplest model is \(a = -kv\), where \(k > 0\) is a drag coefficient with units s⁻¹.

当涉及空气阻力时，加速度往往依赖于速度。最简单的模型是 \(a = -kv\)，其中 \(k > 0\) 是单位为 s⁻¹ 的阻力系数。

Separate the variables:

分离变量：

dv/dt = -kv ⇒ ∫ (1/v) dv = -k∫ dt

Integrating and using \(v = u\) at \(t = 0\) gives:

积分并利用 \(t = 0\) 时 \(v = u\)，得到：

v(t) = u exp(-kt)

Then displacement is obtained by integrating the velocity:

位移通过对速度积分得到：

x(t) = (u/k)(1 – exp(-kt))

As \(t → ∞\), the total distance approaches \(u/k\). This is a classic result for speed-dependent drag.

当 \(t → ∞\) 时，总距离趋近于 \(u/k\)。这是速度相关阻力的经典结论。

5. Acceleration as a Function of Displacement | 加速度为位移的函数

If acceleration is given as a function of \(x\), the direct link between acceleration and velocity is:

如果加速度是 \(x\) 的函数，加速度与速度之间的直接联系是：

a = v dv/dx

Integrating with respect to \(x\):

对 \(x\) 积分：

∫ a dx = ∫ v dv = ½(v² – u²)

This formula is extremely useful when \(a(x)\) is known, because it gives \(v\) as a function of \(x\) without needing time.

当 \(a(x)\) 已知时，这个公式非常有用，因为它能直接给出 \(v\) 关于 \(x\) 的函数，而不需要时间。

For simple harmonic motion, \(a = -ω²x\). If the particle is released from rest at \(x = x₀\), then:

对于简谐运动，\(a = -ω²x\)。若质点从 \(x = x₀\) 处由静止释放，则：

v² = ω²(x₀² – x²)

This equation describes how speed is largest at equilibrium and zero at the extremes.

该方程说明速度在平衡位置最大，在两端为零。

6. Graphical Interpretation | 图形解释

On an acceleration–time graph, the area between the curve and the \(t\)-axis equals the change in velocity.

在加速度-时间图像中，曲线与 \(t\) 轴之间的面积等于速度的变化量。

On a velocity–time graph, the gradient at any point equals the acceleration, and the area under the curve equals the displacement.

在速度-时间图像中，任意一点的斜率等于加速度，曲线下的面积等于位移。

On a displacement–time graph, the gradient equals the velocity, and the curvature shows whether the acceleration is positive or negative.

在位移-时间图像中，斜率等于速度，而弯曲方向表明加速度为正还是为负。

These graphical links remain true even when acceleration is variable.

即使加速度是变化的，这些图形关系仍然成立。

7. Worked Example: Time-Dependent Acceleration | 例题：加速度随时间变化

Example: a particle moves in one dimension with acceleration \(a = 2t + 1\) m/s². At \(t = 0\), \(v = -3\) m/s and \(x = 2\) m. Find \(v(t)\) and \(x(t)\).

例题：一质点沿一维运动，加速度 \(a = 2t + 1\) m/s²。在 \(t = 0\) 时，\(v = -3\) m/s，\(x = 2\) m。求 \(v(t)\) 和 \(x(t)\)。

First integrate the acceleration:

先对加速度积分：

v(t) = ∫ (2t + 1) dt = t² + t + C

Using \(v(0) = -3\) gives \(C = -3\), so:

由 \(v(0) = -3\) 得 \(C = -3\)，所以：

v(t) = t² + t – 3

Now integrate again:

再积分一次：

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One-Dimensional Motion with Variable Acceleration | 一维变加速度运动分析