📚 Equations of Motion | 运动学方程
The equations of motion describe the relationships among displacement, velocity, acceleration and time for objects moving with constant acceleration. In IB Physics, mastering these kinematic equations is essential for analysing linear motion, free fall and projectile problems. They give you a powerful set of tools to predict motion when acceleration is uniform.
运动学方程描述了物体在恒定加速度下位移、速度、加速度和时间之间的关系。在 IB 物理中，掌握这些运动方程对于分析直线运动、自由落体和抛体问题至关重要。当加速度均匀时，这组方程为你提供了预测运动的有力工具。
1. Introduction to Kinematic Equations | 运动学方程简介
In IB Physics, when an object moves in a straight line with constant acceleration, its motion can be described by a set of four equations. These are often called the SUVAT equations, named after the five variables they contain: s (displacement), u (initial velocity), v (final velocity), a (constant acceleration) and t (time). The equations are derived from the definitions of velocity and acceleration and are only valid when acceleration remains unchanged throughout the motion.
在 IB 物理中，当物体以恒定加速度沿直线运动时，其运动可由四个方程描述。这些方程常被称为 SUVAT 方程，以其包含的五个变量命名：s（位移）、u（初速度）、v（末速度）、a（恒定加速度）和 t（时间）。这些方程由速度和加速度的定义推导而来，并且仅当整个运动过程中加速度保持恒定时才成立。
2. The Four Key Equations (SUVAT) | 四个关键方程 (SUVAT)
The four SUVAT equations that you must memorise and apply correctly in IB Physics are:
IB 物理中必须记住并正确应用的四个 SUVAT 方程是：
v = u + at – relates final velocity, initial velocity, acceleration and time.
v = u + at —— 关联末速度、初速度、加速度和时间。
s = ut + ½at²; – calculates displacement when time is known, including the contribution of acceleration.
s = ut + ½at²; —— 在已知时间时计算位移，并计入加速度的贡献。
s = ½(u + v)t – gives displacement using the average velocity when both initial and final velocities are known.
s = ½(u + v)t —— 当已知初速度和末速度时，利用平均速度求位移。
v² = u² + 2as – links velocities and displacement without involving time.
v² = u² + 2as —— 在不需要时间的情况下，将速度和位移联系起来。
3. Equation 1: v = u + at | 方程一：v = u + at
This equation is a direct rearrangement of the definition of acceleration. Since acceleration is the rate of change of velocity, a = (v - u) / t, multiplying both sides by t and adding u gives v = u + at. It tells you how velocity changes under constant acceleration over a given time interval.
该方程由加速度的定义直接变形而来。因为加速度是速度的变化率，即 a = (v - u) / t，两边同时乘以 t 并加上 u 就得到 v = u + at。它告诉你，在恒定加速度下，经过一段时间后速度如何变化。
al velocity is v = 4.0 + 1.5 × 6.0 = 13 m s⁻¹. Always check that your units are consistent – in IB Physics, SI units are expected.
例如，一位自行车手以初速度 u = 4.0 m s⁻¹ 出发，并以 a = 1.5 m s⁻² 加速 t = 6.0 s，末速度 v = 4.0 + 1.5 × 6.0 = 13 m s⁻¹。始终要检查单位一致——IB 物理期望使用国际单位制。
4. Equation 2: s = ut + ½at²; | 方程二：s = ut + ½at²;
This equation calculates the total displacement when an object accelerates uniformly from an initial velocity u over time t. The first term ut represents the displacement if there were no acceleration, and the second term ½at² accounts for the additional displacement caused by acceleration.
该方程计算物体从初速度 u 开始、在时间 t 内匀加速运动的总位移。第一项 ut 表示无加速度时的位移，第二项 ½at² 表示由加速度引起的额外位移。
You can derive it by substituting v = u + at into s = ½(u+v)t erates at a = 2.0 m s⁻² for 5.0 s: s = 10×5.0 + ½×2.0×(5.0)² = 50 + 25 = 75 m.
你可以通过将 v = u + at 代入 s = ½(u+v)t 推导出这个方程。在实际中，当已知 u、a 和 t 而需求 s 时，使用此方程。例如，一辆汽车以 u = 10 m s⁻¹ 运动，以 a = 2.0 m s⁻² 加速 5.0 秒：s = 10×5.0 + ½×2.0×(5.0)² = 50 + 25 = 75 m。
5. Equation 3: s = ½(u + v)t | 方程三：s = ½(u + v)t
When acceleration is constant, the average velocity is simply the mean of the initial and final velocities: average velocity = (u + v) / 2. Displacement is then average velocity multiplied by time. This equation is extremely useful when you know both the starting and ending speeds and the time, but not the acceleration.
当加速度恒定时，平均速度等于初速度和末速度的算术平均值：平均速度 = (u + v) / 2。位移等于平均速度乘以时间。该方程在你已知初速度和末速度以及时间、但不知加速度时极其有用。
30 + 10)×20 = 400 m. You did not need the deceleration value to find the distance covered.
设想一列火车在 t = 20 s 内从 u = 30 m s⁻¹ 匀减速至 v = 10 m s⁻¹。位移 s = ½(30 + 10)×20 = 400 m。你甚至不需要减速度的数值就求出了前进的距离。
6. Equation 4: v² = u² + 2as | 方程四：v² = u² + 2as
This time-independent equation is derived by eliminating t from the first two equations. It links velocities directly to displacement and acceleration. Use it whenever time is not given or not asked for in a problem.
这个不含时间的方程是通过从前两个方程中消去 t 推导出来的。它将速度与位移和加速度直接联系起来。当题目中未给出或未要求时间时，即可使用该方程。
p2; / (2g). The equation is also central to stopping distance calculations in road safety.
例如，当你竖直抛出一个球并需求最大高度时，最高点的末速度 v = 0。已知 u 和 a（a = -g），可直接求解 s：0 = u² + 2(-g)s → s = u² / (2g)。该方程在道路安全的刹车距离计算中也处于核心地位。
7. Derivation from Definitions | 由定义推导
All four equations stem from two fundamental definitions. Acceleration is defined as the rate of change of velocity:
这四个方程均源于两个基本定义。加速度定义为速度的变化率：
a = (v - u) / t → v = u + at
Rearranging gives the first equation. Average velocity for constant acceleration is:
整理后得到第一个方程。恒定加速度下的平均速度为：
vav = (u + v) / 2
Then displacement is s = vav t = ½(u+v)t sup2;. Eliminating t from v = u + at and s = ½(u+v)t gives v² = u² + 2as.
然后位移为 s = vav t = ½(u+v)t，得到第三个方程。将 v = u + at 代入，即得 s = ut + ½at²;。从 v = u + at 和 s = ½(u+v)t 中消去 t，则得到 v² = u² + 2as。
Understanding these derivations strengthens your grasp of the physics and reduces your reliance on blind formula memorisation.
理解这些推导过程能够加深你对物理原理的掌握，并减少对公式死记硬背的依赖。
8. Choosing the Right Equation | 选择合适的方程
In IB Physics problems, you will often be given three known quantities and asked to find a fourth. The key is to identify which SUVAT variable is missing from the statement. Write down the five symbols and tick the ones you know, then select the equation that excludes the unknown quantity you don't need.
在 IB 物理题目中，你通常会得到三个已知量，要求求解第四个量。关键是要找出题目陈述中缺失了哪个 SUVAT 变量。列出这五个符号并勾出已知量，然后选择那个不包含你所不需求解量的方程。
For instance, if a question gives u, v and s and asks for a, time t is not involved – ac12;at². If you struggle, draw a quick table of the five quantities and mark which are known.
例如，若题目给出了 u、v 和 s 并求 a，则与时间 t 无关——应选择 v² = u² + 2as。若已知 u、a 和 t 并需要 s，则使用 s = ut + ½at²;。如果你感觉困难，可以快速画一个包含五个量的表格，标记出已知量。
- Missing a → s = ½(u+v)t
- Missing s → v = u + at (or combine with other equation)
- Missing v → s = ut + ½at²;
- Missing u → s = vt - ½at²; (derived, but rare in IB)
- Missing t → v² = u² + 2as
9. Sign Conventions and Vector Direction | 符号约定与矢量方向
a downward throw, u will also be negative. Sticking to one sign convention throughout the calculation is vital.
位移、速度和加速度都是矢量；在运动方程中其方向至关重要。你必须在每个问题开始时选定一个正方向。对于竖直运动，通常取向上的方向为正。此时重力加速度为 a = -g = -9.8 m s⁻²。对于向下抛掷，u 也会是负值。在整个计算过程中坚持一种符号约定至关重要。
20.4 m. If you had taken downward as positive, u would be negative and a positive, yet the magnitude of displacement would be the same.
一个常见的 IB 考题：球以 20 m s⁻¹ 向上抛出。取向上的方向为正，则 u = +20 m s⁻¹，a = -9.8 m s⁻²。在最高点 v = 0。求最大高度时使用 v² = u² + 2as：0 = (20)² + 2(-9.8)s → s = 20.4 m。如果你以向下为正，u 将为负而 a 为正，但位移的大小相同。
10. Free Fall and Vertical Motion | 自由落体与竖直运动
Free fall is motion under gravity alone, with negligible air resistance. Near the Earth' equations particularly simple: s = ½gt²; and v = gt (if downward is positive).
自由落体是仅受重力作用、空气阻力可忽略的运动。在地表附近，加速度为 a = g = 9.8 m s⁻² 方向向下。运动学方程直接适用，a = ±g 取决于所选的正方向。当物体从静止释放时，u = 0，方程变得特别简单：若以下方为正，s = ½;gt²;，v = gt。
For an object thrown vertically, the motion is symmetrical: the time to rise equals the time to fall back to the original level. Use the equations to find the time of flight, maximum height or impact velocity. Remember to treat upward and downward segments consistently with your sign convention.
对于竖直抛出的物体，其运动具有对称性：上升到最高点的时间等于落回原高度的时间。利用这些方程可求出飞行时间、最大高度或撞击速度。要记住，在上升和下降阶段始终按你的符号约定来处理。
11. Example Problems and Step-by-Step Solutions | 例题与逐步解答
Example 1: Car acceleration
A car accelerates from rest at 3.0 m s⁻² for 8.0 seconds. Calculate the final velocity and the distance covered.
例题1：汽车加速
一辆汽车从静止开始以 3.0 m s⁻² 的加速度行驶 8.0 秒。求末速度和行驶距离。
= 24 m s⁻¹. For distance, s = ut + ½at² = 0 + ½×3.0×(8.0)² = 96 m.
解答：u = 0，a = 3.0 m s⁻²，t = 8.0 s。求末速度，使用 v = u + at：v = 0 + 3.0×8.0 = 24 m s⁻¹。求距离，s = ut + ½at² = 0 + ½×3.0×(8.0)² = 96 m。
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