Reading and Analysing Displacement-Time Graphs | 位移-时间图像的读图与分析

📚 Reading and Analysing Displacement-Time Graphs | 位移-时间图像的读图与分析

In A-Level Mathematics (Mechanics) and Physics, the displacement-time graph is one of the most powerful tools for describing motion. It shows how an object’s position changes relative to a fixed origin over time, and it allows us to read off velocities, identify stationary moments, and distinguish between distance and displacement at a glance.

在 A-Level 数学（力学部分）和物理中，位移-时间图像是描述运动最强大的工具之一。它展示了物体相对于固定原点随时间如何改变位置，使我们能够一目了然地读出速度、识别静止时刻，并区分路程与位移。

1. Understanding the Axes | 理解坐标轴

The horizontal axis (x-axis) always represents time \(t\), measured in seconds (s). The vertical axis (y-axis) represents displacement \(s\), measured in metres (m). Displacement is a vector quantity, meaning it has both magnitude and direction; the sign of \(s\) tells us which side of the origin the object is on.

水平轴（x 轴）始终表示时间 \(t\)，单位为秒（s）。垂直轴（y 轴）表示位移 \(s\)，单位为米（m）。位移是矢量，既有大小又有方向；\(s\) 的正负号告诉我们物体位于原点的哪一侧。

Every point on the graph has coordinates \((t, s)\), meaning “at time \(t\), the object is at displacement \(s\) from the origin”. If \(s\) is positive, the object is on the positive side of the origin; if negative, it is on the negative side.

图上的每一点都有坐标 \((t, s)\)，表示”在时刻 \(t\)，物体距原点位移为 \(s\)”。若 \(s\) 为正，物体在原点正方向一侧；若为负，则在负方向一侧。

2. The Gradient Is Velocity | 斜率即速度

The single most important skill in reading a displacement-time graph is understanding that the gradient (slope) at any point equals the instantaneous velocity of the object. Mathematically, for a straight-line segment between two points \((t_1, s_1)\) and \((t_2, s_2)\):

读位移-时间图像最重要的一项技能是理解：图像上任意一点的斜率等于物体的瞬时速度。数学上，对于连接两点 \((t_1, s_1)\) 和 \((t_2, s_2)\) 的直线段：

velocity \(v = \dfrac{\Delta s}{\Delta t} = \dfrac{s_2 – s_1}{t_2 – t_1}\)

The units of this gradient are metres per second (ms⁻¹), which are exactly the units of velocity.

斜率的单位为米每秒（ms⁻¹），这正是速度的单位。

A positive gradient means the object is moving in the positive direction. A negative gradient means it is moving in the negative direction (towards the origin or beyond it). A zero gradient means the object is stationary.

正斜率表示物体沿正方向运动；负斜率表示物体沿负方向运动（朝向原点或越过原点）；零斜率表示物体静止。

3. Straight Lines: Constant Velocity | 直线段：匀速运动

When the displacement-time graph is a straight line, the gradient is constant, so the velocity is constant. The object is moving with uniform velocity. For example, a line rising steadily at 30° to the horizontal indicates constant positive velocity; the steeper the line, the greater the speed.

当位移-时间图像是直线时，斜率为常数，因此速度恒定，物体做匀速运动。例如，一条以 30° 角平稳上升的直线表示恒定的正向速度；直线越陡，速率越大。

Consider the graph of a car that travels 120 m in 8 s at constant velocity. The gradient is \(120/8 = 15\) ms⁻¹. The line is straight and passes through the origin. If the same car then turns around and returns 60 m in 5 s, the second segment has gradient \((-60)/5 = -12\) ms⁻¹.

设想一辆汽车以恒定速度在 8 秒内行驶 120 m。斜率为 \(120/8 = 15\) ms⁻¹，该直线为过原点的直线。若之后汽车掉头，在 5 秒内返回 60 m，则第二段斜率为 \((-60)/5 = -12\) ms⁻¹。

Notice that the steepness of the second segment (magnitude 12) is less than the first (magnitude 15), meaning the car returns more slowly than it set out.

注意第二段的陡峭程度（大小为 12）小于第一段（大小为 15），说明汽车返回时比出发时慢。

4. Curved Graphs: Changing Velocity | 曲线：变速运动

If the graph curves, the gradient is changing, so the velocity is changing — the object is accelerating or decelerating. The instantaneous velocity at a particular time is the gradient of the tangent drawn to the curve at that point.

如果图像是弯曲的，斜率在变化，因此速度也在变化——物体正在加速或减速。某一时刻的瞬时速度等于该时刻曲线切线的斜率。

When the curve becomes steeper over time (concave upward), the velocity is increasing: the object is speeding up. When the curve becomes flatter over time (concave downward), the velocity is decreasing: the object is slowing down.

当曲线随时间变得越来越陡（凹向上），速度在增大：物体在加速。当曲线随时间变得越来越平缓（凹向下），速度在减小：物体在减速。

It is essential to draw the tangent accurately using a ruler — A-Level examiners award marks for the method: draw a tangent, form a right-angled triangle, and compute rise over run.

准确画切线至关重要——A-Level 考官会根据方法给分：画切线、构造直角三角形、计算纵坐标增量除以横坐标增量。

5. Distance vs. Displacement | 路程与位移的区别

The graph always records displacement, not distance. Suppose an object moves 30 m in the positive direction, then 20 m back towards the origin. Its final displacement is 10 m, but the total distance travelled is 50 m.

图像记录的是位移，不是路程。假设物体先向正方向移动 30 m，再向原点方向返回 20 m。其最终位移为 10 m，但总路程为 50 m。

On the graph, displacement is read directly from the vertical coordinate. Topologically, the vertical height of the final point gives the final displacement; the total distance must be calculated by adding the absolute changes in displacement over each segment of the motion.

在图像上，位移直接从纵坐标读出。从图形角度看，终点的纵坐标高度给出最终位移；而总路程必须通过将各运动段位移变化量的绝对值相加来计算。

For example, a graph that goes from \(s = 0\) to \(s = 30\), then back to \(s = 10\), has a total distance of \(|30 – 0| + |10 – 30| = 30 + 20 = 50\) m, while the net displacement is simply \(10\) m.

例如，图像从 \(s = 0\) 到 \(s = 30\)，再回到 \(s = 10\)，总路程为 \(|30 – 0| + |10 – 30| = 30 + 20 = 50\) m，而净位移仅为 \(10\) m。

6. Average Velocity and Average Speed | 平均速度与平均速率

Average velocity is defined as total displacement divided by total time. Average speed is defined as total distance divided by total time. These two quantities are equal only when the object never changes direction.

平均速度定义为总位移除以总时间；平均速率定义为总路程除以总时间。只有当物体从不改变方向时，这两个量才相等。

average velocity = \(\dfrac{\text{final displacement} – \text{initial displacement}}{\text{total time}}\)

average speed = \(\dfrac{\text{total distance}}{\text{total time}}\)

On a displacement-time graph, the average velocity over interval \([t_1, t_2]\) equals the gradient of the chord (straight line) connecting the two corresponding points on the graph. The average speed, however, requires knowing the path taken, which the graph does not always directly reveal if the object doubles back.

在位移-时间图像上，区间 \([t_1, t_2]\) 上的平均速度等于连接图上两个对应点的弦（直线）的斜率。然而，平均速率需要知道实际路径，如果物体折返，图像并不总能直接显示。

7. Turning Points and Stationary Instants | 转折点与瞬时静止

When the graph reaches a local maximum or minimum, the gradient is momentarily zero. This is a turning point: the object stops instantaneously before reversing direction. At these points, the velocity is 0 ms⁻¹ but the displacement is not necessarily zero.

当图像达到局部最大值或最小值时，斜率为瞬时零，这是转折点：物体在反向之前瞬时静止。在这些点上，速度为 0 ms⁻¹，但位移不一定为零。

For instance, a ball thrown vertically upwards reaches its highest point where the displacement is maximum; on the graph this appears as a smooth peak. The tangent at the peak is horizontal, confirming zero velocity.

例如，竖直上抛的球到达最高点时位移最大；在图像上表现为平滑的峰值。峰值处的切线是水平的，证实速度为零。

Conversely, a horizontal segment of the graph (not just a point) indicates the object is at rest for an extended period of time. The gradient is zero throughout that interval. This is common in journey graphs that include rest stops.

相反，图像上的一段水平线段（不仅仅是一个点）表示物体在一段时间内保持静止，整个区间内斜率为零。这在包含中途休息的行程图中很常见。

8. Worked Example: Reading a Multi-Stage Journey | 典型例题：解读多阶段行程

Consider the following graph describing a cyclist’s journey:

考虑以下描述骑行者行程的图像：

Stage A: (0 s, 0 m) to (10 s, 100 m) — straight line.

阶段 A：(0 s, 0 m) 到 (10 s, 100 m)——直线。
Stage B: (10 s, 100 m) to (25 s, 100 m) — horizontal.

阶段 B：(10 s, 100 m) 到 (25 s, 100 m)——水平线。
Stage C: (25 s, 100 m) to (40 s, 40 m) — straight, negative gradient.

阶段 C：(25 s, 100 m) 到 (40 s, 40 m)——负斜率直线。

During Stage A, the velocity is \((100 – 0)/(10 – 0) = 10\) ms⁻¹ (positive, moving away from origin). During Stage B, the velocity is 0 ms⁻¹ — the cyclist is resting. During Stage C, the velocity is \((40 – 100)/(40 – 25) = -4\) ms⁻¹ (negative, moving back towards the origin).

在阶段 A，速度为 \((100 – 0)/(10 – 0) = 10\) ms⁻¹（正值，远离原点运动）。在阶段 B，速度为 0 ms⁻¹——骑行者正在休息。在阶段 C，速度为 \((40 – 100)/(40 – 25) = -4\) ms⁻¹（负值，向原点返回）。

The total displacement is \(40 – 0 = 40\) m. The total distance is \(|100| + |40 – 100| = 100 + 60 = 160\) m. The total time is 40 s. Therefore average velocity \(= 40/40 = 1\) ms⁻¹, while average speed \(= 160/40 = 4\) ms⁻¹.

总位移为 \(40 – 0 = 40\) m。总路程为 \(|100| + |40 – 100| = 100 + 60 = 160\) m。总时间为 40 s。因此平均速度为 \(40/40 = 1\) ms⁻¹，而平均速率为 \(160/40 = 4\) ms⁻¹。

9. Sketching Graphs from a Description | 根据文字描述画图

A common exam task is to sketch a displacement-time graph from a verbal description of motion. To do this correctly, work step by step:

常见的考试题型是根据文字描述画出位移-时间图像，为此应逐步处理：

Identify the starting displacement: does the motion begin at the origin or elsewhere?

确定初始位移：运动是从原点开始还是从别处开始？
Determine whether each stage has constant velocity (straight line), acceleration (curve), or rest (horizontal line).

判断每个阶段是匀速（直线）、加速（曲线）还是静止（水平线）。
Check the direction: does the displacement increase, decrease, or stay the same?

检查方向：位移是增大、减小还是保持不变？
Ensure continuity — the object cannot teleport; the graph must be a continuous curve or line.

确保连续性——物体不能瞬移；图像必须是一条连续曲线或直线。

Also remember: if the object decelerates but never reverses direction, the graph curves towards the horizontal but never crosses it. Only when the direction changes does the gradient change sign.

同时记住：如果物体减速但从未反向，图像会向水平方向弯曲但不会穿过它。只有当方向改变时，斜率才会改变正负号。

10. Common Mistakes and Exam Advice | 常见错误与备考建议

Students frequently confuse displacement-time graphs with velocity-time graphs. Remember: in a displacement-time graph, the gradient is velocity; in a velocity-time graph, the area under the curve is displacement. Never apply the area rule to a displacement-time graph.

学生经常将位移-时间图像与速度-时间图像混淆。请记住：在位移-时间图中，斜率是速度；在速度-时间图中，曲线下的面积是位移。切勿将面积法则应用于位移-时间图像。

Another common error is mistaking “returning to the origin” for “the final displacement being zero” — this is correct only if the object ends at the origin. Also, a point where the graph crosses the time axis means the object passes through the origin, not that it is stationary.

另一个常见错误是将”回到原点”误认为”最终位移为零”——只有当物体最终位于原点时才成立。此外，图像与时间轴的交点意味着物体经过原点，而不是静止。

When drawing tangents, use a sharp pencil and a transparent ruler; make the triangle as large as possible to minimise percentage error. Always state the units (ms⁻¹) in your final answer.

画切线时，请使用削尖的铅笔和透明直尺；三角形尽量做大，以减小百分比误差。最终答案中始终注明单位（ms⁻¹）。

11. Practice Question | 练习题

A particle moves along a straight line. Its displacement-time graph consists of Two parts: first, a straight segment from (0 s, −5 m) to (4 s, 15 m); second, a straight segment from (4 s, 15 m) to (10 s, −3 m).

一质点沿直线运动。其位移-时间图像由两段组成：第一段为从 (0 s, −5 m) 到 (4 s, 15 m) 的直线段；第二段为从 (4 s, 15 m) 到 (10 s, −3 m) 的直线段。

Calculate: (a) the velocity of the first segment; (b) the velocity of the second segment; (c) the total distance travelled; (d) the average speed.

请计算：(a) 第一段的速度；(b) 第二段的速度；(c) 总路程；(d) 平均速率。

Solution: (a) \(v_1 = (15 – (-5))/(4 – 0) = 20/4 = 5\) ms⁻¹. (b) \(v_2 = ((-3) – 15)/(10 – 4) = (-18)/6 = -3\) ms⁻¹. (c) Displacement change first segment \(|15 – (-5)| = 20\) m; second segment \(|-3 – 15| = 18\) m; total distance \(= 38\) m. (d) Total time \(= 10\) s, so average speed \(= 38/10 = 3.8\) ms⁻¹.

解答：(a) \(v_1 = (15 – (-5))/(4 – 0) = 20/4 = 5\) ms⁻¹。(b) \(v_2 = ((-3) – 15)/(10 – 4) = (-18)/6 = -3\) ms⁻¹。(c) 第一段位移变化量 \(|15 – (-5)| = 20\) m；第二段位移变化量 \(|-3 – 15| = 18\) m；总路程 \(= 38\) m。(d) 总时间 \(= 10\) s，因此平均速率 \(= 38/10 = 3.8\) ms⁻¹。

Mastering displacement-time graphs gives you a solid foundation for understanding kinematics — and it is a frequent source of A-Level exam marks. By learning to read gradients, recognise turning points, and distinguish distance from displacement, you will be well equipped to handle any graph-based mechanics question.

掌握位移-时间图像，为理解运动学打下了坚实基础——这也是 A-Level 考试中高频得分的来源。通过学会读取斜率、识别转折点、区分路程与位移，你将能够从容应对任何基于图像的力学问题。

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Reading and Analysing Displacement-Time Graphs | 位移-时间图像的读图与分析