A-Level Physics: Deriving Displacement from Graphs | 由图像推导位移

📚 A-Level Physics: Deriving Displacement from Graphs | 由图像推导位移

In kinematics, graphical analysis is one of the most powerful tools to understand motion. By reading or calculating the area under different motion graphs, we can directly determine displacement. This article explains the methods and pitfalls of deriving displacement from displacement–time, velocity–time, and acceleration–time graphs, with worked examples aligned to the CIE A-Level syllabus.

在运动学中，图像分析是理解运动最有效的工具之一。通过读取或计算不同运动图像下的面积，我们可以直接确定位移。本文将解释从位移-时间、速度-时间和加速度-时间图像推导位移的方法和常见陷阱，并结合 CIE A-Level 考纲给出例题。

1. The displacement–time graph | 位移-时间图像

The most direct way to find displacement is from a displacement–time (s–t) graph. On this graph, the vertical axis gives the displacement from a reference point at any instant. For example, if a particle starts at s = 0 m and moves to s = 5 m, the graph is simply a straight horizontal line at 5 m after the motion stops. To find the displacement at a specific time, you read the value of s directly from the curve.

直接从位移-时间（s–t）图像上获取位移是最直接的方法。该图像中纵轴表示任意时刻相对于参考点的位移。例如，若质点从 s = 0 m 运动到 s = 5 m，且之后静止，图像就是一条高度为 5 m 的水平直线。要找到某时刻的位移，只需从曲线上读出对应的 s 值即可。

Additionally, the gradient of the s–t graph gives the velocity. A straight line means constant velocity; a curved line means the velocity changes. Thus, while the s–t graph gives displacement directly, it is not necessary to integrate anything to obtain the displacement — it is already displayed.

此外，s–t 图像的斜率给出速度。直线表示速度恒定；曲线表示速度变化。因此，虽然 s–t 图像直接显示位移，但不需要任何积分运算——位移已经展示在图中。

2. The velocity–time graph and area | 速度-时间图像与面积

For a velocity–time (v–t) graph, displacement is not read directly from the y-axis. Instead, the area between the graph and the time axis represents the displacement. This arises from the fundamental relationship v = ds/dt, so s = ∫ v dt. In graphical terms, integrating the velocity over time is equivalent to finding the area under the v–t curve.

对于速度-时间（v–t）图像，位移不能直接从纵轴读取。相反，图像与时间轴之间的面积表示位移。这源于基本关系 v = ds/dt，因此 s = ∫ v dt。从图像角度看，对速度在时间上积分就等于求 v–t 曲线下的面积。

When the velocity is constant, the graph is a horizontal line, and the area is simply the rectangle v × t. When velocity changes uniformly, the graph is a straight line with a constant slope, and the area can be calculated using the area of a trapezium or a triangle. For non‑uniform acceleration, the curve is not linear, and numerical methods or counting squares are required.

当速度恒定时，图像是一条水平线，面积为矩形，计算式为 v × t。当速度均匀变化时，图像是一条斜率恒定的直线，面积可用梯形或三角形面积公式计算。对于非匀加速度，曲线不是直线，此时需要数值方法或数方格。

3. Calculating areas: rectangle, triangle, trapezium | 计算面积：矩形、三角形、梯形

The simplest calculations of displacement from a v–t graph use geometric formulas. For a constant velocity v over a time interval t, the displacement is:

从 v–t 图像计算位移的最简单方法是使用几何公式。若在时间 t 内速度恒为 v，则位移为：

s = v × t

If the velocity changes uniformly from u to v over time t, the graph is a straight line, and the shape under it is a trapezium. Its area equals the average of the two velocities multiplied by the time:

若速度在时间 t 内从 u 均匀变化到 v，图像是一条直线，其下的形状为梯形。其面积等于两速度的平均值乘以时间：

s = ½ (u + v) × t

For a body starting from rest and reaching a velocity v in time t, the graph is a triangle, and the displacement becomes s = ½ v t. These formulas match the kinematic equations exactly.

对于从静止出发并在时间 t 内达到速度 v 的物体，图像为三角形，位移变为 s = ½ v t。这些公式与运动学方程完全吻合。

4. Counting squares and strip approximation | 数方格与条带近似

When the v–t graph is curved, the area cannot be found by simple shapes. In an examination, the graph is often drawn on graph paper, and you can count the number of small squares under the curve. Each square has an area equal to (unit of v) × (unit of t). Multiplying the total number of squares by that unit area gives the displacement.

当 v–t 图像是曲线时，面积无法用简单形状求出。在考试中，图像通常画在坐标纸上，你可以数出曲线下小方格的个数。每个方格的面积等于 (速度单位) × (时间单位)。将总方格数乘以该单位面积即可得到位移。

Alternatively, you can split the area into vertical strips of equal width Δt. Approximate each strip as a rectangle whose height equals the average velocity in that strip, or as a trapezium. Summing the areas of all strips yields an approximate total displacement. The smaller the strip width, the more accurate the approximation; this is the basis of numerical integration.

另一种方法是将区域分割成等宽 Δt 的垂直条带。将每个条带近似为高度为该条带平均速度的矩形，或近似为梯形。将所有条带面积求和得到近似总位移。条带越窄，近似越精确；这就是数值积分的基本原理。

5. Deriving s = ut + ½at² from the v–t graph | 从 v–t 图像推导 s = ut + ½at²

One of the most elegant uses of a v–t graph is deriving the second equation of motion. Consider a body with initial velocity u and constant acceleration a. The graph is a straight line with slope a, starting at u. The area under the graph from t = 0 to t = t consists of a rectangle of area ut (the velocity if it remained constant) plus a triangle of area ½ × t × (at) = ½at². Therefore:

v–t 图像最优雅的应用之一就是推导运动学第二方程。设物体初速度为 u，加速度恒为 a。图像是一条斜率为 a 的直线，从 u 开始。从 t=0 到 t=t 的曲线下面积由一个矩形（面积 ut，对应速度不变的情况）和一个三角形（面积 ½ × t × (at) = ½at²）组成。因此：

s = ut + ½at²

This graphical derivation clearly shows that the term ut accounts for the displacement at constant initial velocity, while ½at² accounts for the extra displacement caused by the changing velocity. In the CIE syllabus, you are expected to recall and apply this equation, but understanding its origin strengthens your conceptual grasp.

这一图形推导清晰地表明，ut 项对应初速度恒定时产生的位移，而 ½at² 项则是由速度变化导致的额外位移。在 CIE 考纲中，你需要记住并会应用这个方程，但理解其来源能加深你的概念理解。

6. Non-uniform acceleration: graphical integration | 非匀加速度：图像积分

When acceleration is not constant, the v–t graph is not a straight line. However, the area under the curve always equals the displacement, because displacement is defined as the time integral of velocity. This holds true for any curve. To evaluate the area, you may use counting squares or the strip method, as described earlier, or if the function v(t) is known, you can integrate analytically.

当加速度不是常数时，v–t 图像不是直线。然而，曲线下的面积始终等于位移，因为位移定义为速度对时间的积分。这适用于任何曲线。为了计算面积，可以使用前面介绍的数方格或条带法，或者如果已知 v(t) 函数，也可以直接解析积分。

For example, if v = t² m/s, the displacement between t = 1 s and t = 3 s is given by ∫₁³ t² dt = [t³/3]₁³ = (27 − 1)/3 = 8.67 m. In a graph, this corresponds to the area of the region bounded by the curve, the time axis, and the vertical lines t = 1 and t = 3.

例如，若 v = t² m/s，则 t=1 s 到 t=3 s 之间的位移为 ∫₁³ t² dt = [t³/3]₁³ = (27 − 1)/3 = 8.67 m。在图像上，这对应于由曲线、时间轴以及 t=1 和 t=3 两条垂线所围成的区域面积。

7. Acceleration–time graphs: a two‑step process | 加速度-时间图像：两步过程

An acceleration–time (a–t) graph does not directly give displacement. The area under an a–t graph gives the change in velocity (Δv = ∫ a dt). To extract displacement, you first need to construct a v–t graph from the a–t data, then find the area under that v–t graph. If you only have an a–t graph, you must integrate twice or use the kinematic equations if the acceleration is constant.

加速度-时间（a–t）图像不能直接给出位移。a–t 图像下的面积给出速度变化量（Δv = ∫ a dt）。要提取位移，你首先需要根据 a–t 数据构建 v–t 图像，然后再求 v–t 图像下的面积。如果你只有 a–t 图像，就必须二次积分，或者当加速度为常数时使用运动学方程。

Consider a body at rest at t = 0, with acceleration a = 2 m/s² for 5 s, then a = 0 for the next 3 s. The a–t graph shows a rectangle of height 2 from t=0 to t=5. Area = 2 × 5 = 10 m/s, so the velocity at t=5 is 10 m/s. The v–t graph then has a straight line from 0 to 10 m/s over the first 5 s, then a horizontal line at 10 m/s for the next 3 s. The displacement after 8 s is the area under the v–t graph: trapezium area = ½ × (10 + 10) × 3 + ½ × 5 × 10 = 30 + 25 = 55 m.

考虑一个在 t=0 时静止的物体，前 5 s 内加速度 a=2 m/s²，之后 3 s 内 a=0。a–t 图像在 t=0 到 t=5 之间是一个高为 2 的矩形。面积 = 2 × 5 = 10 m/s，因此 t=5 时速度为 10 m/s。v–t 图像在前 5 s 是一条从 0 到 10 m/s 的直线，之后 3 s 是高度为 10 m/s 的水平线。8 s 后的位移就是 v–t 图像下的面积：梯形面积 = ½ × (10 + 10) × 3 + ½ × 5 × 10 = 30 + 25 = 55 m。

8. Worked example: displacement from a v–t graph | 例题：从 v–t 图像求位移

Let’s apply these methods to a typical exam question. A particle moves along a straight line. Its velocity–time graph is shown in Figure 1 (described here). The graph is a straight line from (0 s, 4 m/s) to (6 s, 10 m/s), followed by a horizontal line from (6 s, 10 m/s) to (10 s, 10 m/s). Calculate the total displacement after 10 s.

让我们将这些方法应用于一道典型考试题。质点沿直线运动，其速度-时间图像如图1所示（此处文字描述）。图像是一条从 (0 s, 4 m/s) 到 (6 s, 10 m/s) 的直线，接着是一条从 (6 s, 10 m/s) 到 (10 s, 10 m/s) 的水平线。求 10 s 后的总位移。

For the first 6 s, the shape under the graph is a trapezium. The area is s₁ = ½ × (4 + 10) × 6 = 42 m. For the next 4 s, the shape is a rectangle: s₂ = 10 × 4 = 40 m. Total displacement s = 82 m. If the graph had shown a negative velocity below the time axis, the area between the curve and the time axis would be treated as negative displacement, meaning motion in the opposite direction.

前 6 s 内，图像下的形状为梯形。面积 s₁ = ½ × (4 + 10) × 6 = 42 m。接下来 4 s 为矩形：s₂ = 10 × 4 = 40 m。总位移 s = 82 m。如果图像在时间轴下方出现速度为负的区域，则曲线与时间轴之间的面积应视为负位移，表示反方向运动。

9. Common mistakes: distance vs. displacement | 常见错误：路程与位移的区别

A frequent error is confusing total distance with displacement when part of the v–t graph lies below the time axis. For displacement, you must subtract the area below the axis from the area above the axis. For distance, you add the absolute values of the areas. For example, if a body moves forward 5 m and then backward 3 m, the displacement is 2 m, but the distance travelled is 8 m.

一个常见错误是当 v–t 图像部分位于时间轴下方时，将总路程与位移混淆。对于位移，必须用轴上方面积减去轴下方面积；对于路程，则将各块面积的绝对值相加。例如，若物体先前进 5 m 再后退 3 m，位移为 2 m，但路程为 8 m。

Another pitfall is using the gradient of a v–t graph to find displacement. The gradient gives acceleration, not displacement. Displacement is found from the area, not the slope. Always check which quantity is on the y‑axis: if it is velocity, use area; if it is displacement, use the y‑value directly.

另一个陷阱是用 v–t 图像的斜率来求位移。斜率给出的是加速度，不是位移。位移来自面积，而非斜率。始终检查纵轴是什么：如果是速度，就用面积；如果是位移，就直接读取纵轴值。

10. Summary: what each graph gives | 小结：每种图像能给出什么

The table below summarizes how to extract displacement and other useful quantities from motion graphs. In all cases, the graph must be read carefully to identify which quantity is plotted on each axis.

下表总结了如何从运动图像中提取位移及其他有用量。在任何情况下，都要仔细读取图像，确认哪个量绘制在哪个轴上。

Graph	Gradient gives	Area gives
Displacement–time	Velocity	Not used (displacement on y‑axis)
Velocity–time	Acceleration	Displacement
Acceleration–time	Rate of change of acceleration	Change in velocity

For a velocity–time graph, remember that “area under the graph” always refers to the area between the curve and the time axis, with areas below the axis counted as negative for displacement. For a curved v–t graph, use counting squares or strip methods to approximate the area. In an exam, always include the correct units (metres) and a clear statement of your method.

对于速度-时间图像，记住“图像下的面积”总是指曲线与时间轴之间的面积，轴下方的面积在计算位移时计为负值。对于弯曲的 v–t 图像，使用数方格或条带法来近似面积。在考试中，务必写出正确单位（米）以及清晰的解题步骤。

11. Final tips for exams | 考试冲刺建议

When answering questions on deriving displacement from graphs, follow these steps: 1) Identify the type of graph. 2) Decide whether you need the y‑value (s–t graph) or the area (v–t graph). 3) Split the area into simple shapes or count squares for curved graphs. 4) Apply the appropriate formula, being careful with signs. 5) State the answer with units and a meaningful direction if required.

在回答有关由图像推导位移的问题时，请按以下步骤：1) 确认图像类型；2) 判断是需要读取纵轴值（s–t 图）还是面积（v–t 图）；3) 将面积分割成简单形状，或对于曲线采用数方格；4) 套用相应公式并注意正负号；5) 给出带单位的答案，必要时说明方向。

Practise with past CIE papers, where such questions frequently appear in Paper 2 and Paper 4. With time, you will quickly recognise whether a question tests gradient or area, and you will solve it correctly.

利用 CIE 真题进行练习，这类问题常见于 Paper 2 和 Paper 4。假以时日，你就能迅速判断题目考查的是斜率还是面积，并准确解答。

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A-Level Physics: Deriving Displacement from Graphs | 由图像推导位移