Deducing Displacement | 位移的推导

📚 Deducing Displacement | 位移的推导

Displacement is one of the most fundamental concepts in kinematics, yet students often confuse it with distance. In CIE A-Level Physics, deducing displacement means finding the change in position of an object from graphs, equations, or experimental data. This article explains the key methods and common pitfalls.

位移是运动学中最基本的概念之一，但学生常常将它与路程混淆。在 CIE A-Level 物理中，推导位移意味着从图像、方程或实验数据中求出物体位置的变化。本文将解释主要方法和常见误区。

1. Displacement vs Distance | 位移与距离

Displacement is a vector quantity: it has both magnitude and direction. Distance is a scalar quantity: it only measures how much ground an object has covered along its path.

位移是矢量：既有大小又有方向。路程是标量：它只测量物体沿路径经过的总长度。

If a runner completes one lap of a 400 m track, the distance travelled is 400 m, but the displacement is zero because the start and end points coincide.

如果一名跑步者绕 400 m 跑道跑完一圈，路程为 400 m，但位移为零，因为起点和终点重合。

In one-dimensional motion, displacement can be positive or negative depending on the chosen coordinate direction, whereas distance is always positive.

在一维运动中，位移可为正或负，取决于所选的坐标方向，而路程始终为正。

2. Defining Displacement from Position Vectors | 由位置矢量定义位移

Displacement s is defined as the change in position: s = x − x₀ in one dimension, or s = r − r₀ in two or three dimensions.

位移 s 定义为位置的变化：一维中 s = x − x₀，二维或三维中 s = r − r₀。

Because displacement is a vector, a negative value in one-dimensional motion simply means the object has moved in the negative coordinate direction.

由于位移是矢量，一维运动中的负值仅仅表示物体沿坐标负方向运动。

s = Δx = x − x₀
s = Δr = r − r₀

上述方程表明，位移只取决于初末位置，与中间路径无关。

3. Displacement from Velocity-Time Graphs | 从速度-时间图求位移

The area between a velocity-time graph and the time axis gives the displacement. Areas above the time axis are positive displacement; areas below the axis are negative displacement.

速度-时间图与时间轴之间的面积表示位移。时间轴上方的面积为正位移；时间轴下方的面积为负位移。

For a graph with both positive and negative regions, add the signed areas algebraically to obtain the net displacement.

对于同时具有正、负区域的图像，应代数相加各带符号面积，以求得净位移。

If you instead add the absolute values of the areas, you obtain the total distance travelled, not displacement.

如果你改为将各面积的绝对值相加，得到的是总路程，而不是位移。

When the velocity-time graph is curved, you can still deduce displacement by counting squares under the curve or by dividing the area into narrow trapeziums and summing their areas.

当速度-时间图为曲线时，仍然可以通过计数曲线下方的方格或将面积分割成窄梯形并求和来推导位移。

This is an estimate because the curve may not follow simple straight-line shapes exactly. The smaller the strips, the better the approximation.

这是一种估算，因为曲线可能并不完全符合简单的直线形状。条带越窄，近似越好。

4. Uniform Velocity and the Area Rule | 匀速运动与面积法则

When velocity is constant at v, the velocity-time graph is a horizontal line. The area under the graph over time t is a rectangle of height v and width t, so displacement is s = vt.

当速度恒为 v 时，速度-时间图是一条水平线。在时间 t 内图像下方的面积是一个高为 v、宽为 t 的矩形，因此位移为 s = vt。

This is the simplest case of deducing displacement from a graph, and it also shows why the SI unit of displacement is the metre, obtained from m s⁻¹ × s = m.

这是从图像推导位移的最简单情形，它也说明位移的 SI 单位是米，由 m s⁻¹ × s = m 得到。

If the velocity is negative, the graph lies below the time axis, and the area gives a negative displacement in the chosen positive direction.

如果速度为负，图像位于时间轴下方，该面积给出沿选定正方向的负位移。

5. Uniform Acceleration: The suvat Equations | 匀加速运动：suvat 方程

For constant acceleration a, initial velocity u, final velocity v, and time t, the displacement can be deduced using the suvat equations.

对于恒定加速度 a、初速度 u、末速度 v 和时间 t，位移可用 suvat 方程推导。

s = ut + ½at²

This equation comes from the area under a velocity-time graph: a rectangle of area ut plus a triangle of area ½at².

该方程来自速度-时间图下方的面积：一个面积为 ut 的矩形加上一个面积为 ½at² 的三角形。

v² = u² + 2as

This form is useful when the time t is unknown, allowing displacement to be deduced from initial and final speeds.

当时间 t 未知时，这种形式很有用，可以从初速度和末速度推导位移。

Also, using average velocity: s = ((u + v)/2) × t, which is valid only when acceleration is constant.

此外，利用平均速度：s = ((u + v)/2) × t，但仅在加速度恒定时有效。

6. Deducing Displacement from Acceleration-Time Graphs | 从加速度-时间图推导位移

An acceleration-time graph does not directly give displacement. First, the area under an acceleration-time graph gives the change in velocity, Δv.

加速度-时间图不能直接给出位移。首先，加速度-时间图下方的面积给出速度变化量 Δv。

To deduce displacement, you need the initial velocity u. Then find the velocity at any time by adding the area to u, and plot or integrate the velocity-time graph to find the area.

要推导位移，需要知道初速度 u。然后将面积加到 u 上求出任意时刻的速度，再绘制或求速度-时间图下方的面积。

Step 1: Δv = area under a-t graph
Step 2: v(t) = u + Δv(t)
Step 3: s = area under v-t graph

第 1 步：Δv = a-t 图下方面积；第 2 步：v(t) = u + Δv(t)；第 3 步：s = v-t 图下方面积。

For example, if u = 2 m s⁻¹ and the acceleration is a constant 5 m s⁻² for 4 s, then Δv = 20 m s⁻¹, so v = 22 m s⁻¹. With constant acceleration, the displacement is s = ((2 + 22)/2) × 4 = 48 m.

例如，若 u = 2 m s⁻¹，加速度恒为 5 m s⁻² 持续 4 s，则 Δv = 20 m s⁻¹，因此 v = 22 m s⁻¹。由于加速度恒定，位移为 s = ((2 + 22)/2) × 4 = 48 m。

7. Free Fall and Vertical Displacement | 自由落体与竖直位移

For an object falling freely near Earth’s surface, the acceleration is g = 9.81 m s⁻² downwards. Choosing upward as positive gives a = −g.

对于在地球表面附近自由下落的物体，加速度为向下的 g = 9.81 m s⁻²。若取向上为正，则 a = −g。

If an object is dropped from rest at height h, its displacement after time t is s = ut + ½at² = 0 − ½gt². The negative sign means downward displacement.

如果物体从高度 h 处由静止释放，经过时间 t 后的位移为 s = ut + ½at² = 0 − ½gt²。负号表示位移向下。

The final position can then be found from y = h + s. If the object falls for enough time to reach the ground, setting y = 0 allows you to calculate the fall time.

末位置可由 y = h + s 求得。如果物体下落足够长时间到达地面，令 y = 0 即可计算下落时间。

If the object returns to its starting height, the total vertical displacement is zero, even though the distance travelled is not.

如果物体回到初始高度，总的竖直位移为零，尽管路程不为零。

8. Projectile Motion: Combining Components | 抛体运动：分量的合成

In projectile motion, horizontal displacement x and vertical displacement y are independent. The horizontal motion has constant velocity, and the vertical motion has constant acceleration g downwards.

在抛体运动中，水平位移 x 和竖直位移 y 相互独立。水平方向为匀速运动，竖直方向为向下的匀加速运动。

For a projectile launched with speed u at angle θ to the horizontal:

对于以速率 u、与水平方向成 θ 角发射的抛体：

x = u cos θ × t

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Deducing Displacement | 位移的推导