📚 Solving Problems Involving Velocity and Time Using Vectors | 利用向量解决速度与时间问题
In A-Level Mechanics, vectors provide a powerful tool for describing motion in two dimensions. By expressing position, velocity, and acceleration as vectors, we can solve complex problems involving direction and magnitude. This article covers key techniques for solving problems where velocity and time are linked through vectors, following the Edexcel specification. We will explore constant velocity, constant acceleration, relative motion, closest approach, and interception, all using vector methods.
在 A-Level 力学中，向量为描述二维运动提供了强大的工具。通过将位置、速度和加速度表示为向量，我们可以解决涉及方向与大小（模）的复杂问题。本文遵循 Edexcel 考试大纲，涵盖利用向量联系速度与时间的关键解题技巧。我们将探讨匀速、匀加速、相对运动、最近距离以及拦截问题，全部使用向量方法。
1. Vector Representation of Motion | 运动的向量表示
In two dimensions, the position vector r of a particle is given by r = x i + y j, where x and y are functions of time t. The velocity vector v is the rate of change of displacement: v = dr/dt = (dx/dt) i + (dy/dt) j. Similarly, acceleration a = dv/dt = d²r/dt². The magnitude of velocity is speed: |v| = √((dx/dt)² + (dy/dt)²).
在二维空间中，粒子的位置向量 r 由 r = x i + y j 给出，其中 x 和 y 是时间 t 的函数。速度向量 v 是位移的变化率：v = dr/dt = (dx/dt) i + (dy/dt) j。类似地，加速度 a = dv/dt = d²r/dt²。速度的大小即为速率：|v| = √((dx/dt)² + (dy/dt)²)。
Every component is treated independently. The i-direction is horizontal and the j-direction is vertical (unless otherwise stated). This separation allows us to use scalar equations for each axis while keeping the overall direction information within the vector form.
每个分量都独立处理。i 方向通常为水平方向，j 方向为竖直方向（除非另有说明）。这种分离使我们能够对每个轴使用标量方程，同时在向量形式中保留整体方向信息。
2. Position, Velocity, and Acceleration Vectors | 位置、速度和加速度向量
To find velocity from position, differentiate each component with respect to t. For example, if r = (t² + 3t) i + (2t − 1) j, then v = (2t + 3) i + 2 j and a , given acceleration and initial conditions, integrate to obtain velocity and position. If a = 4 i − 2 j and initially v = 3 i + j at t = 0, then v(t) = ∫a dt = (4t + 3) i + (−2t + 1) j.
要从位置求速度，对每个分量关于 t 求导。例如，若 r = (t² + 3t) i + (2t − 1) j，则 v = (2t + 3) i + 2 j，a = 2 i + 0 j。反之，给定加速度和初始条件，通过积分得到速度和位置。若 a = 4 i − 2 j 且在 t = 0 时 v = 3 i + j，则 v(t) = ∫a dt = (4t + 3) i + (−2t + 1) j。
Further integration gives the position vector r(t), using the initial position r₀. For instance, if r₀ = i − 2 j, then r(t) = ∫v + 3t + 1) i + (−t² + t − 2) j. These operations underpin most vector kinematics problems.
再次积分，利用初始位置 r₀，可得位置向量 r(t)。例如，若 r₀ = i − 2 j，则 r(t) = ∫v dt = (2t² + 3t + 1) i + (−t² + t − 2) j。这些运算是大多数向量运动学问题的基础。
3. Constant Velocity Problems | 匀速问题
When a particle moves with constant velocity v, its position vector at time t is r = r₀ + v t, where r / time. The direction of motion is constant, and both components of velocity remain unchanged.
当粒子以恒定速度 v 运动时，其在时间 t 的位置向量为 r = r₀ + v t，其中 r₀ 为初始位置。这是一维速度公式的向量推广。运动方向恒定，且速度的两个分量均保持不变。
Example: A boat starts at position 2 i + 3 j (in km) and travels with constant velocity 5 i − 2 j km/h. After 4 hours, its position is r j. We can also ask: at what time does the boat cross the line where the i-coordinate is 30? Set 2 + 5t = 30, giving t = 5.6 hours, then find the j-coordinate: 3 − 2×5.6 = -8.2.
举例：一艘小船从位置 2 i + 3 j（公里）出发，以恒定速度 5 i − 2 j 公里/小时航行。4 小时后，位置为 r = (2 + 5×4) i + (3 − 2×4) j = 22 i − 5 j。我们也可以问：小船何时穿过 i 坐标为 30 的直线？令 2 + 5t = 30，得 t = 5.6 小时，然后计算 j 坐标：3 − 2×5.6 = -8.2。
4. Constant Acceleration and SUVAT in Vector Form | 匀加速及向量形式的SUVAT
The familiar SUVAT equations extend naturally to vectors: v = u + a t, s = u t + ½ a t², and s = v t − ½ a t². Here s is the displacement vector, u initial velocity, v final velocity, and a ctor dot product, which is rarely required in Edexcel A-Level, but you may use the magnitude form with caution when motions are collinear.
我们熟悉的 SUVAT 方程可以自然地推广到向量：v = u + a t，s = u t + ½ a t²，以及 s = v t − ½ a t²。这里 s 是位移向量，u 为初速度，v 为末速度，a 为恒定加速度。请注意：标量方程 v² = u² + 2as 转换为向量点乘，在 Edexcel A-Level 中较少考察，但在共线运动下可谨慎使用其模长形式。
v = u + a t
s = u t + ½ a t²
Example: A particle accelerates from u = 2 i + 3 j with a = 4 i − 2 j. After 3 s, v = (2 + 12) i + (3 − 6) j = 14 i − 3 j. The displacement is s 9 j + 18 i − 9 j = 24 i. So the particle is 24 units along the i-direction from its start.
举例：一粒子从 u = 2 i + 3 j 出发，加速度 a = 4 i − 2 j。3 秒后，v = (2 + 12) i + (3 − 6) j = 14 i − 3 j。位移 s = (2 i + 3 j)×3 + ½ (4 i − 2 j)×9 = 6 i + 9 j + 18 i − 9 j = 24 i。因此粒子在 i 方向上距离起点 24 个单位。
5. Finding Time When Two Objects Meet | 求两物体相遇的时间
To determine if and when two moving particles collide, we set their position vectors equal at the same time t: r₁(t) = r₂(t). This gives two scalar equations (for i and j components). Solve for t from both; the solutions must be consistent for a collision to occur.
要判断两个运动粒子是否及何时相撞，令它们在相同时刻 t 的位置向量相等：r₁(t) = r₂(t)。这给出两个标量方程（i 和 j 分量）。分别求解 t，两者必须一致才发生碰撞。
Example: Particle A has rₐ = (3t + 1) i + (2t − 4) j, particle B has r₆ = (t + 7) i + (t + 2) j. Equating i-components: 3t + 1 = t + 7 ⇒ t = 3. Equating j-components: 2t − 4 = t + 2 ⇒ t = 6. Times differ, so the particles do not collide – they pass through the same i-position at different moments, but never meet simultaneously.
举例：粒子 A 的位置为 rₐ = (3t + 1) i + (2t − 4) j，粒子 B 的位置为 r₆ = (t + 7) i + (t + 2) j。令 i 分量相等：3t + 1 = t + 7 ⇒ t = 3。令 j 分量相等：2t − 4 = t + 2 ⇒ t = 6。时间不同，因此粒子不会相撞——它们在不同时刻经过同一 i 位置，但从不真正相遇。
A common exam question gives one particle an unknown constant velocity and asks you to find the velocity required for interception. You would introduce an unknown vector v te both position equations, and solve the components for a and b so that the same t works.
常见考题会给其中一个粒子一个未知的恒定速度，要求求出拦截所需的速度。你需设未知向量 v = a i + b j，写出双方的位置方程，并解出 a 和 b 使得同一个 t 成立。
6. Calculating the Distance Between Two Moving Objects | 计算两运动物体间的距离
The vector from object A to B is d = r₆ − rₐ. The distance between them at time t is the magnitude D(t) = |d| = √xd² + yd², where xd and yd are the i and j components of d. To find the distance at a specific time, simply substitute t.
从物体 A 到 B 的向量为 d = r₆ − rₐ。它们在时间 t 的距离为其模长 D(t) = |d| = √xd² + yd²，其中 xd 和 yd 是 d 的 i 分量和 j 分量。要求特定时刻的距离，直接代入 t 即可。
akes a minimum value, it is simpler to work with the square of the distance, f(t) = D² = xd² + yd², which is a quadratic in t. Differentiate f(t) with respect to t, set f'(t) = 0, and solve for t. Then verify it is a minimum by checking f''(t) > 0 or by completing the square.
要求距离取最小值的时间，更简便的方法是使用距离的平方 f(t) = D² = xd² + yd²，它是关于 t 的二次函数。对 f(t) 关于 t 求导，令 f'(t) = 0，解出 t。然后通过检查 f''(t) > 0 或配方法来确认其为最小值。
Example: A moves as rₐ = (2t) i + (t+1) j, B moves as r₆ = (3t−4) i + (−t+5) j. Then d 4) i + (−2t + 4) j. f(t) = (t−4)² + (−2t+4)² = t² −8t +16 + 4t² −16t+16 = 5t² −24t +32. f'(t) = 10t −24
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