Deducing Displacement | 推导位移

📚 Deducing Displacement | 推导位移

Displacement is the vector quantity that describes the change in position of an object. It is defined as the straight-line distance from the starting point to the finishing point, along with the direction. In kinematics, being able to deduce displacement from given information — such as initial velocity, acceleration, time, or graphical data — is a core skill for A‑Level Physics students. This article explores the various methods used to deduce displacement, including graphical analysis, algebraic derivation of the SUVAT equations, and the principles of integration.

位移是描述物体位置变化的矢量。它被定义为从起点到终点的直线距离及其方向。在运动学中，根据给定的信息（如初速度、加速度、时间或图形数据）推导位移是A-Level物理学生的核心技能。本文探讨了推导位移的各种方法，包括图形分析、匀加速运动方程的代数推导以及积分原理。

1. What is Displacement? | 什么是位移？

Displacement is not the same as distance. Distance is a scalar quantity that measures the total path length travelled, while displacement only considers the net change in position. For example, if an object moves 5 m east and then 3 m west, the total distance is 8 m, but the displacement is 2 m east. The SI unit for displacement is the metre (m).

位移不同于路程。路程是标量，测量经过的路径总长度，而位移只考虑位置的净变化。例如，一个物体向东移动5米，然后向西移动3米，总路程是8米，但位移是向东2米。位移的国际单位是米（m）。

In one‑dimensional motion along a straight line, displacement can be calculated as Δx = x − x₀, where x is the final position and x₀ is the initial position. The sign indicates direction.

在沿直线的一维运动中，位移可以按 Δx = x − x₀ 计算，其中 x 是末位置，x₀ 是初位置。符号表示方向。

2. Displacement from a Velocity–Time Graph | 从速度–时间图求位移

One of the most powerful tools for deducing displacement is the velocity–time (v–t) graph. The area enclosed between the velocity graph and the time axis represents displacement. This is because area = velocity × time, which has the dimensions of displacement.

从速度–时间图（v–t图）推导位移是最有效的方法之一。速度曲线与时间轴之间所围成的面积表示位移。这是因为面积 = 速度 × 时间，量纲与位移一致。

For uniform motion where velocity is constant, the v–t graph is a horizontal line. The displacement is simply the area of the rectangle: s = v × t.

在速度恒定的匀速运动中，v–t图是一条水平线。位移就是矩形的面积：s = v × t。

For uniformly accelerated motion, the graph is a sloping straight line. The area under such a graph is a trapezium. If the initial velocity is u and the final velocity is v after time t, the area (displacement) is given by s = ½(u + v)t.

对于匀加速运动，图形是一条倾斜的直线。该图下的面积是一个梯形。如果初速度为 u，经过时间 t 后末速度为 v，则面积（位移）为 s = ½(u + v)t。

3. Area Under the Curve: Geometric Methods | 曲线下的面积：几何方法

When the v–t graph forms a more complex shape, you can still deduce displacement by dividing the area into standard geometric figures such as rectangles, triangles and trapeziums. The total displacement is the sum of the areas of these shapes, taking care to treat areas below the time axis as negative displacement (indicating motion in the opposite direction).

当v–t图形成更复杂的形状时，你仍然可以通过将面积分割成矩形、三角形和梯形等标准几何图形来推导位移。总位移是这些图形面积之和，注意时间轴以下的面积应视为负位移（表示反向运动）。

For example, during a journey where a car speeds up, travels at a constant speed, then slows down, the v–t graph consists of a triangle, a rectangle and another triangle. The net displacement is the sum of these three areas.

例如，在一次汽车先加速、再匀速、后减速的行程中，v–t图由一个三角形、一个矩形和另一个三角形组成。净位移就是这三个面积之和。

Area of triangle = ½ × base × height

三角形面积 = ½ × 底 × 高
Area of rectangle = base × height

矩形面积 = 底 × 高
Area of trapezium = ½ × (sum of parallel sides) × height

梯形面积 = ½ × (上底 + 下底) × 高

4. Deriving the SUVAT Equation s = ut + ½at² | 推导匀加速运动方程 s = ut + ½at²

The equation s = ut + ½at² is one of the fundamental SUVAT equations for uniformly accelerated motion. It can be deduced directly from a velocity–time graph. For constant acceleration a, the final velocity after time t is v = u + at. The displacement is the area under the v–t graph, which is the area of a trapezium with parallel sides u and v, and width t.

方程 s = ut + ½at² 是匀加速运动的基本SUVAT方程之一。它可以直接从速度–时间图推导出来。对于恒定加速度 a，经过时间 t 的末速度为 v = u + at。位移是v–t图下的面积，即上底 u、下底 v、宽 t 的梯形面积。

Starting with s = ½(u + v)t and substituting v = u + at gives:

由 s = ½(u + v)t 出发，代入 v = u + at 得到：

s = ½(u + u + at)t = ½(2u + at)t = ut + ½at²

This derivation shows that the term ut represents the displacement if the object had continued at the initial velocity, and ½at² is the additional displacement due to acceleration.

这一推导表明，ut 项表示若物体以初速度继续运动所产生的位移，而 ½at² 是由于加速度而产生的附加位移。

5. Using Average Velocity | 使用平均速度

For uniformly accelerated motion starting with velocity u and ending with velocity v, the average velocity vₐᵥ is defined as (u + v)/2. Displacement can then be expressed simply as s = vₐᵥ × t. This approach is particularly useful when the initial and final velocities are known, but the acceleration is not explicitly given.

对于初速度为 u、末速度为 v 的匀加速运动，平均速度 vₐᵥ 定义为 (u + v)/2。那么位移可以简单地表示为 s = vₐᵥ × t。当已知初速度和末速度而加速度未明确给出时，这种方法尤其有用。

Be careful: the formula s = vₐᵥ t using (u+v)/2 applies only when acceleration is constant. For non‑uniform acceleration, the average velocity is not simply the arithmetic mean unless the acceleration changes linearly with time, which is rare in simple problems.

请注意：使用 (u+v)/2 的 s = vₐᵥ t 公式仅适用于加速度恒定的情况。对于非匀加速，除非加速度随时间线性变化（这在简单问题中很少见），否则平均速度不能简单取算术平均值。

6. Deriving v² = u² + 2as | 推导 v² = u² + 2as

Another essential SUVAT equation relates final velocity, initial velocity, acceleration and displacement without involving time. This equation can be deduced by eliminating t from the two equations v = u + at and s = ut + ½at².

另一个重要的SUVAT方程将末速度、初速度、加速度和位移联系起来，而不涉及时间。可以通过从 v = u + at 和 s = ut + ½at² 中消去 t 来推导该方程。

From v = u + at, we obtain t = (v − u)/a. Substituting this into the displacement equation:

由 v = u + at 得 t = (v − u)/a。将其代入位移方程：

s = u × (v − u)/a + ½a × ((v − u)/a)²

s = (uv − u²)/a + ½a × (v² − 2uv + u²)/a²

s = (uv − u²)/a + (v² − 2uv + u²)/(2a)

Multiply through by 2a: 2as = 2(uv − u²) + v² − 2uv + u²

2as = v² − u², hence v² = u² + 2as

This equation is very practical for deducing displacement when time is unknown, for example in braking distance calculations.

当时间未知时，此方程在推导位移方面非常实用，例如在刹车距离计算中。

7. Displacement from a Non‑Uniform Acceleration | 非匀加速运动中的位移

When acceleration is not constant, the simple area method using triangles and rectangles may not be sufficient. However, if the velocity function v(t) is known, displacement can be found by summing the area of very narrow strips under the curve — essentially performing integration. In graph work without a known function, one can estimate displacement by counting squares or using the trapezium rule.

当加速度不恒定时，使用三角形和矩形的简单面积法可能不够。但如果已知速度函数 v(t)，则可以通过对曲线下极窄条带面积求和——本质上是进行积分——来求得位移。在没有已知函数的情况下，可以通过数格或使用梯形法则来估算位移。

The trapezium rule approximates the area by dividing the time interval into equal segments, calculating the area of each segment as (v₁ + v₂)/2 × Δt, and summing them. The more segments used, the better the approximation.

梯形法则通过将时间区间等分，计算每段面积为 (v₁ + v₂)/2 × Δt 并求和来近似面积。段数越多，近似越精确。

8. Integration Approach: s = ∫ v dt | 积分法：s = ∫ v dt

In A‑Level Physics, you may encounter situations where you are given an equation for velocity as a function of time, such as v = 5 + 2t³. Displacement is then the definite integral of velocity with respect to time between the limits of the time interval: s = ∫ v dt from t₁ to t₂.

在A-Level物理中，你可能会遇到给出速度作为时间函数的方程的情况，例如 v = 5 + 2t³。那么位移就是速度对时间的定积分，积分限为时间区间的起点和终点：s = ∫ v dt 从 t₁ 到 t₂。

For the example v = 5 + 2t³ (m/s) from t = 0 to t = 2 s:

以 v = 5 + 2t³ (m/s) 从 t = 0 到 t = 2 秒为例：

s = ∫₀² (5 + 2t³) dt = [5t + (2t⁴)/4]₀² = [5t + ½t⁴]₀² = (10 + 8) − 0 = 18 m

This integration method is a direct extension of the area‑under‑graph concept and applies even when acceleration changes with time in a non‑linear way.

这种积分方法是图形下面积概念的直接推广，即使加速度随时间非线性变化也适用。

9. Displacement in Two Dimensions | 二维位移

In two‑dimensional motion, displacement is a vector with both x‑ and y‑components. To deduce the overall displacement, you must treat the horizontal and vertical motions independently. For projectile motion, the horizontal displacement is calculated using constant horizontal velocity, while the vertical displacement uses SUVAT equations with acceleration g = 9.81 m/s² downwards.

在二维运动中，位移是一个具有 x 分量和 y 分量的矢量。要推导总位移，必须独立处理水平和垂直运动。对于抛体运动，水平位移使用恒定的水平速度计算，而垂直位移使用以向下 g = 9.81 m/s² 为加速度的SUVAT方程计算。

The magnitude of the resultant displacement is given by |s| = √(sₓ² + sᵧ²), and the direction is given by the angle θ = tan⁻¹(sᵧ / sₓ).

合位移的大小由 |s| = √(sₓ² + sᵧ²) 给出，方向由角 θ = tan⁻¹(sᵧ / sₓ) 给出。

For instance, a ball kicked horizontally at 10 m/s off a cliff takes 2 s to hit the ground. Horizontal displacement sₓ = 10 × 2 = 20 m. Vertical displacement sᵧ = ½ × 9.81 × 2² = 19.6 m. The resultant displacement is √(20² + 19.6²) ≈ 28 m at an angle of about 44° below the horizontal.

例如，一个以10 m/s水平踢出悬崖的球，经过2秒落地。水平位移 sₓ = 10 × 2 = 20 m。垂直位移 sᵧ = ½ × 9.81 × 2² = 19.6 m。合位移为 √(20² + 19.6²) ≈ 28 m，方向与水平面夹角约44°向下。

10. Worked Example: Braking Distance | 例题：刹车距离

A car travelling at 20 m/s decelerates uniformly at 5 m/s² until it stops. Deduce the displacement (braking distance).

一辆以20 m/s行驶的汽车以5 m/s²的加速度匀减速直至停止。请推导位移（刹车距离）。

Using v² = u² + 2as, where u = 20 m/s, v = 0 m/s, a = −5 m/s² (deceleration is negative acceleration). Rearranging: 0 = 20² + 2(−5)s → 0 = 400 − 10s → s = 40 m. The braking distance is 40 m.

使用 v² = u² + 2as，其中 u = 20 m/s，v = 0 m/s，a = −5 m/s²（减速即负加速度）。整理得：0 = 20² + 2(−5)s → 0 = 400 − 10s → s = 40 m。刹车距离为40米。

Alternatively, using s = ½(u+v)t, first find t: v = u + at → 0 = 20 − 5t → t = 4 s. Then s = ½(20+0)×4 = 40 m. Both methods yield the same result.

或者使用 s = ½(u+v)t，先求 t：v = u + at → 0 = 20 − 5t → t = 4 s。然后 s = ½(20+0)×4 = 40 m。两种方法结果相同。

11. Common Pitfalls | 常见误区

One typical mistake is confusing displacement with distance. Always check whether the velocity graph crosses the time axis. If it does, the total distance travelled is the sum of the absolute values of the individual areas, whereas displacement is the net signed area.

一个典型错误是将位移与路程混淆。务必检查速度图是否穿过时间轴。如果穿过，总路程是各面积绝对值之和，而位移是带符号的净面积。

Another error is applying SUVAT equations to motion with non‑constant acceleration. Remember that the SUVAT equations are only valid when acceleration is uniform. If acceleration varies, you must use graphical methods or integration.

另一个错误是将SUVAT方程应用于加速度不恒定的运动。请记住，SUVAT方程仅在加速度均匀时有效。若加速度变化，必须使用图形法或积分。

Also, when deducing displacement in two dimensions, do not add the x and y components without vector resolution. Always use Pythagoras’ theorem for the magnitude and trigonometry for the direction.

此外，在推导二维位移时，不要在没有矢量分解的情况下直接相加 x 和 y 分量。务必使用勾股定理求大小和三角学求方向。

12. Summary | 总结

Deducing displacement is a central theme in kinematics, and there are multiple pathways to find it. From the geometric area under a velocity–time graph to algebraic manipulation of SUVAT equations, and from the integration of a velocity function to vector analysis in two dimensions, the method chosen depends on the data available. Practising these techniques builds a robust foundation for tackling more complex mechanics problems in A‑Level Physics.

推导位移是运动学的核心主题，且有多种求法。从速度–时间图下的几何面积，到SUVAT方程的代数变换，从速度函数的积分到二维矢量分析，所选方法取决于已知数据。通过练习这些方法，你将为解决A-Level物理中更复杂的力学问题打下坚实基础。

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Deducing Displacement | 推导位移