Velocity-Time Graph Analysis | 速度-时间图像解析

📚 Velocity-Time Graph Analysis | 速度-时间图像解析

A velocity-time (v-t) graph is one of the most powerful visual tools in A-Level mechanics. It shows how the velocity of a moving object changes over time, and from it, we can read the acceleration, displacement, and distance travelled directly. Mastering v-t graphs is essential for solving kinematics problems efficiently.

速度-时间（v-t）图像是 A-Level 力学中最强大的可视化工具之一。它展示运动物体速度随时间的变化情况，从中我们可以直接读取加速度、位移和运动路程。熟练掌握 v-t 图像是高效解决运动学问题的关键。

1. The Gradient — Acceleration | 斜率——加速度

The gradient of a velocity-time graph at any instant gives the acceleration at that instant. Mathematically, acceleration a = dv/dt, which is exactly the slope of the tangent to the curve at a given time. A positive gradient means positive acceleration; a negative gradient means deceleration (or negative acceleration).

速度-时间图像上任意时刻的斜率代表该时刻的瞬时加速度。数学上，加速度 a = dv/dt，即曲线上某点切线的斜率。斜率为正表示加速，斜率为负表示减速（负加速度）。

For a straight-line segment, the gradient is constant, so the acceleration is uniform. The steeper the line, the greater the magnitude of acceleration.

对于直线段，斜率为常数，因此加速度是恒定的。直线越陡，加速度的绝对值越大。

2. The Area — Displacement | 面积——位移

The area between a velocity-time graph and the time axis (t-axis) represents the displacement of the object. Since displacement = velocity × time for constant velocity, the area of each small strip (v × Δt) summed over the interval gives the total displacement. This is equivalent to evaluating ∫v dt over the time interval.

速度-时间图像与时间轴（t轴）之间的面积代表物体的位移。由于恒定速度下位移 = 速度 × 时间，将每个小条面积（v × Δt）在区间内求和即得总位移。这等价于在时间区间上计算 ∫v dt。

It is crucial to treat areas above the t-axis as positive displacement and areas below the t-axis as negative displacement. The net displacement is the algebraic sum, while total distance is the sum of absolute areas.

必须注意：t轴上方的面积计为正位移，t轴下方的面积计为负位移。净位移是代数和，而总路程是各段面积绝对值之和。

3. Constant Velocity | 恒定速度

When an object moves with constant velocity, its v-t graph is a horizontal straight line. The gradient is zero, so acceleration is zero. The displacement over a time interval Δt is simply v × Δt, which equals the area of the rectangle under the line.

当物体以恒定速度运动时，其 v-t 图像是一条水平直线。斜率为零，因此加速度为零。在时间间隔 Δt 内，位移就是 v × Δt，等于直线下方矩形的面积。

s = v × Δt (units: m)

This is the simplest case, but it forms the foundation for understanding more complex graphs where velocity changes.

这是最简单的情形，但它是理解更复杂速度变化图像的基础。

4. Uniform Acceleration | 匀加速运动

Uniform acceleration produces a straight-line v-t graph with a non-zero slope. The gradient of this line equals the constant acceleration a. The equation of the line is v = u + at, where u is the initial velocity and v is the velocity at time t.

匀加速运动产生一条斜率非零的直线 v-t 图像。该直线的斜率等于恒定加速度 a。直线方程为 v = u + at，其中 u 是初速度，v 是 t 时刻的速度。

The area under a uniform-acceleration line is a trapezium. Its area gives the displacement formula s = ut + ½at². Equivalently, the average velocity (u + v)/2 multiplied by time t gives the same displacement.

匀加速直线下方的面积是一个梯形。其面积给出位移公式 s = ut + ½at²。等价地，平均速度 (u + v)/2 乘以时间 t 也得到相同的位移。

s = (u + v)/2 × t = ut + ½at²

When the line crosses the t-axis, the object momentarily comes to rest and changes direction. The gradient remains the same, but the sign of velocity flips.

当直线与 t 轴相交时，物体瞬间静止并改变方向。斜率保持不变，但速度的符号发生翻转。

5. Non-Uniform Acceleration | 变加速运动

When acceleration is not constant, the v-t graph is a curve rather than a straight line. The gradient of the tangent at any point gives the instantaneous acceleration, which changes continuously with time.

当加速度不恒定时，v-t 图像是一条曲线而非直线。曲线上任意一点切线的斜率给出瞬时加速度，该加速度随时间连续变化。

Common exam curves include v = t², v = √t, and v = sin t. In such cases, integration is required to find displacement, and differentiation is required to find acceleration. Recall:

常见的考试曲线包括 v = t²、v = √t 和 v = sin t。在这种情况下，需要积分求位移、微分求加速度。回忆：

a = dv/dt, s = ∫v dt

For example, if v = t² – 4t + 3, then a = 2t – 4, which changes sign at t = 2. The displacement from t = 0 to t = T is ∫₀ᵀ (t² – 4t + 3) dt = T³/3 – 2T² + 3T.

例如，若 v = t² – 4t + 3，则 a = 2t – 4，加速度在 t = 2 时变号。从 t = 0 到 t = T 的位移为 ∫₀ᵀ (t² – 4t + 3) dt = T³/3 – 2T² + 3T。

6. Distance vs Displacement | 路程与位移

A frequent source of confusion is the difference between total distance travelled and net displacement. Displacement considers direction — it is the integral of velocity. Distance is the integral of speed, i.e., the absolute value of velocity. On a v-t graph, displacement is the signed area (positive above axis, negative below), while distance is the sum of all absolute areas.

一个常见的混淆点是总路程与净位移的区别。位移考虑方向——它是速度的积分。路程是速率的积分，即速度的绝对值。在 v-t 图像上，位移是带符号面积（轴上方为正、下方为负），而路程是所有面积绝对值之和。

To calculate total distance when the graph crosses the t-axis:

Find the time(s) when v = 0 (roots of the velocity equation).
Split the interval at these times.
Integrate |v| separately over each sub-interval and sum the results.

找出 v = 0 的时刻（速度方程的根）。
在这些时刻将区间分段。
在每个子区间上分别对 |v| 积分，再求和。

For example, with v = t² – 4t + 3 = (t – 1)(t – 3), the particle changes direction at t = 1 and t = 3. The total distance in the first 4 seconds is ∫₀¹v dt + (-∫₁³v dt) + ∫₃⁴v dt = 4/3 + 4/3 + 4/3 = 4 m, while the net displacement is only 4/3 m.

例如，v = t² – 4t + 3 = (t – 1)(t – 3)，物体在 t = 1 和 t = 3 时改变方向。前 4 秒的总路程为 ∫₀¹v dt + (-∫₁³v dt) + ∫₃⁴v dt = 4/3 + 4/3 + 4/3 = 4 m，而净位移仅为 4/3 m。

7. Sketching Velocity-Time Graphs | 绘制速度-时间图像

Constructing accurate v-t graphs from given information is a common exam skill. Follow these steps:

Identify axes: time (t) on the horizontal axis, velocity (v) on the vertical axis.
Plot the initial velocity u at t = 0.
For constant acceleration, draw a straight line of slope a.
For changing acceleration, sketch a curve whose gradient matches the given acceleration trend.

确定坐标轴：横轴为时间 t，纵轴为速度 v。
在 t = 0 处标出初速度 u。
匀加速时，画一条斜率为 a 的直线。
变加速时，画出切线斜率符合给定加速度变化趋势的曲线。

Pay attention to key points: intercepts with the axes, turning points where the graph crosses the t-axis, and endpoints. A straight line that returns to the t-axis indicates the object returns to its starting point — the total signed area is zero.

注意关键点：与坐标轴的交点、图像穿过 t 轴的转折点以及端点。如果直线回到 t 轴，表明物体回到出发点——总带符号面积为零。

8. Relating v-t Graphs to Other Graphs | 与 s-t、a-t 图像的关联

Velocity-time graphs connect directly to displacement-time (s-t) and acceleration-time (a-t) graphs. Differentiating s-t gives v-t; differentiating v-t gives a-t. Conversely, integrating a-t gives v-t; integrating v-t gives s-t.

速度-时间图像与位移-时间（s-t）图像和加速度-时间（a-t）图像直接关联。对 s-t 微分得到 v-t；对 v-t 微分得到 a-t。反之，对 a-t 积分得到 v-t；对 v-t 积分得到 s-t。

s-t graph	Gradient → velocity
v-t graph	Gradient → acceleration; Area → displacement
a-t graph	Area → change in velocity

9. Worked Example | 典型例题

Problem: A particle moves in a straight line with velocity v = t² – 6t + 8 for t ≥ 0 (units of m/s and s). Find: (a) the times when the particle is instantaneously at rest; (b) the net displacement from t = 0 to t = 5; (c) the total distance travelled from t = 0 to t = 5.

问题：一质点在直线上运动，速度 v = t² – 6t + 8（单位：m/s 和 s），t ≥ 0。求：(a) 质点瞬时静止的时刻；(b) t = 0 到 t = 5 的净位移；(c) t = 0 到 t = 5 的总路程。

Solution:

解答：

(a) Set v = 0: t² – 6t + 8 = 0 → (t – 2)(t – 4) = 0 → t = 2 s and t = 4 s.

(a) 令 v = 0：t² – 6t + 8 = 0 → (t – 2)(t – 4) = 0 → t = 2 s 和 t = 4 s。

(b) Net displacement s = ∫₀⁵ (t² – 6t + 8) dt = [t³/3 – 3t² + 8t]₀⁵ = (125/3 – 75 + 40) = 125/3 – 35 = 20/3 ≈ 6.67 m.

(b) 净位移 s = ∫₀⁵ (t² – 6t + 8) dt = [t³/3 – 3t² + 8t]₀⁵ = (125/3 – 75 + 40) = 125/3 – 35 = 20/3 ≈ 6.67 m。

(c) Split at t = 2 and t = 4. In [0,2], v ≥ 0: area = ∫₀² (t² – 6t + 8) dt = 8/3 – 12 + 16 = 20/3. In [2,4], v ≤ 0: area = -∫₂⁴ (t² – 6t + 8) dt = -((64/3 – 48 + 32) – (8/3 – 12 + 16)) = -(16/3 – 20/3) = 4/3. In [4,5], v ≥ 0: area = ∫₄⁵ (t² – 6t + 8) dt = (125/3 – 75 + 40) – (64/3 – 48 + 32) = 20/3 – 16/3 = 4/3. Total distance = 20/3 + 4/3 + 4/3 = 28/3 ≈ 9.33 m.

(c) 在 t = 2 和 t = 4 处分段。在 [0,2] 上，v ≥ 0：面积 = ∫₀² (t² – 6t + 8) dt = 8/3 – 12 + 16 = 20/3。在 [2,4] 上，v ≤ 0：面积 = -∫₂⁴ (t² – 6t + 8) dt = -((64/3 – 48 + 32) – (8/3 – 12 + 16)) = -(16/3 – 20/3) = 4/3。在 [4,5] 上，v ≥ 0：面积 = ∫₄⁵ (t² – 6t + 8) dt = (125/3 – 75 + 40) – (64/3 – 48 + 32) = 20/3 – 16/3 = 4/3。总路程 = 20/3 + 4/3 + 4/3 = 28/3 ≈ 9.33 m。

Notice that total distance (9.33 m) exceeds net displacement (6.67 m) because the particle reverses direction twice.

注意总路程（9.33 m）大于净位移（6.67 m），因为质点两次反向运动。

10. Common Pitfalls | 常见误区

Pitfall 1: Confusing displacement with distance. Always check whether an area lies above or below the t-axis before adding it to the total.

误区一：混淆位移与路程。在将面积计入总量前，务必判断该面积位于 t 轴上方还是下方。

Pitfall 2: Forgetting to set v = 0 when finding turning points. The particle only changes direction when velocity equals zero.

误区二：求转折点时忘记令 v = 0。只有当速度为零时，质点才会改变方向。

Pitfall 3: Treating a curved v-t graph as straight in area calculations. If the graph is curved, you must integrate the velocity function, not use trapezium area formulas.

误区三：在面积计算中把弯曲的 v-t 图像当作直线。如果图像是弯曲的，必须对速度函数积分，而不能使用梯形面积公式。

Pitfall 4: Reading gradient incorrectly when the axes have different scales. Always compute the gradient as (Δv)/(Δt) using the actual axis values.

误区四：当两轴刻度不同时读错斜率。务必用实际坐标值计算斜率 (Δv)/(Δt)。

11. Exam Tips | 考试技巧

One of the most reliable strategies is to annotate the graph as you work. Shade the areas you have already calculated, mark the axis-crossing times clearly, and write the velocity function beside the curve. This prevents arithmetic errors and makes your reasoning visible to the examiner.

最可靠的策略之一是边做题边在图上标注。把你已计算过的区域涂上阴影，清楚标记与坐标轴相交的时间，并在曲线旁写出速度函数。这样可以避免算术错误，也能让阅卷人看到你的思路。

Always state the units in your final answer: m/s for velocity, m/s² for acceleration, and m for displacement or distance. Check whether the question asks for displacement or distance before answering.

最终答案务必带上单位：速度用 m/s，加速度用 m/s²，位移或路程用 m。答题前先确认题目问的是位移还是路程。

Memorise the key relationships: gradient gives acceleration, area gives displacement. When in doubt, return to these two fundamental interpretations and everything else will follow.

牢记两个核心关系：斜率给出加速度，面积给出位移。有疑问时，回到这两个最基本的解释，其余问题便会迎刃而解。

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Velocity-Time Graph Analysis | 速度-时间图像解析