Constant Acceleration Formulae 2: Multi-Stage Motion and Gravity | 恒定加速度公式 2：多阶段运动与重力

📚 Constant Acceleration Formulae 2: Multi-Stage Motion and Gravity | 恒定加速度公式 2：多阶段运动与重力

In Mechanics 1, constant acceleration formulae, often called SUVAT equations, are powerful tools for solving problems where acceleration does not change. This second layer of study focuses on multi-stage journeys, vertical motion under gravity, sign conventions, and interpreting velocity-time graphs.

在力学 1 中，恒定加速度公式（常称 SUVAT 方程）是求解加速度不变问题的有力工具。第二层学习重点是多阶段运动、竖直方向的重力运动、符号约定以及速度-时间图的解读。

1. The Five SUVAT Equations | 五个 SUVAT 方程

The five equations link initial velocity u, final velocity v, constant acceleration a, displacement s, and time t.

五个方程将初速度 u、末速度 v、恒定加速度 a、位移 s 和时间 t 联系起来。

v = u + at

s = ut + ½at²

s = (u + v)/2 × t

v² = u² + 2as

s = vt − ½at²

Each equation omits one variable, so choose the equation that does not contain the unknown quantity you are trying to avoid.

每个方程都省略一个变量，因此要选择不含你想避开未知量的那个方程。

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2. Choosing a Positive Direction | 选择正方向

Always state a positive direction before substituting values. For horizontal motion, take “right” or “direction of initial motion” as positive. For vertical motion, take “up” or “down” as positive and keep it throughout the calculation.

代入数值前务必先确定正方向。水平运动通常取 “向右” 或 “初速度方向” 为正；竖直运动取 “向上” 或 “向下” 为正，并在整个计算过程中保持一致。

If up is positive, then gravitational acceleration is a = −9.8 m s⁻². If down is positive, then a = +9.8 m s⁻².

若取向上为正，重力加速度为 a = −9.8 m s⁻²；若取向下为正，则 a = +9.8 m s⁻²。

Displacement, velocity, and acceleration are vectors, so their signs must all follow the chosen convention.

位移、速度和加速度都是矢量，它们的正负号必须都遵循所选的方向约定。

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3. Vertical Motion Under Gravity | 竖直方向的重力运动

Near Earth’s surface, a freely moving projectile has constant acceleration g = 9.8 m s⁻² downwards. Air resistance is ignored in A-Level Mechanics 1 calculations.

在地球表面附近，自由运动的抛体具有向下的大小为 g = 9.8 m s⁻² 的恒定加速度。A-Level 力学 1 的计算通常忽略空气阻力。

For an object thrown upwards, it decelerates on the way up, stops momentarily at the highest point, then accelerates downwards. The velocity at the highest point is v = 0 m s⁻¹.

对于上抛物体，它上升时减速，在最高点瞬间静止，然后向下加速。最高点速度 v = 0 m s⁻¹。

The time to rise to maximum height can be found from v = u + at with v = 0, giving t = u/g when up is positive.

上升到最高点的时间可由 v = u + at 代入 v = 0 求得，取向上为正时 t = u/g。

When an object returns to its original height, the time taken to rise equals the time taken to fall back, as long as air resistance is ignored.

当物体回到原来高度时，只要忽略空气阻力，上升时间等于下落时间。

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4. Multi-Stage Motion: Split the Journey | 多阶段运动：分段处理

Many exam problems combine two or more stages with different accelerations. You must split the motion at the point where acceleration changes, and apply SUVAT separately to each stage.

许多考试题会将两个或多个不同加速度的阶段组合在一起。必须在加速度改变处将运动分段，并对每一段分别应用 SUVAT 方程。

The final velocity of one stage becomes the initial velocity of the next stage only if the acceleration changes instantaneously at that point.

只有当加速度在该点瞬间改变时，前一段的末速度才会成为下一段的初速度。

Record known values for each stage in a table with columns for s, u, v, a, t. This prevents sign errors and shows clearly which variable is missing.

将每一段的已知量记录在表格中，列包括 s、u、v、a、t。这样可以避免符号错误，并清楚显示缺少哪个变量。

Do not carry displacement from one stage into the next unless the question is asking for total displacement over the whole journey.

不要把上一段的位移直接带入下一段，除非题目要求的是整段运动的总位移。

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5. Worked Example: Stone Thrown Upwards | 例题：竖直上抛石块

A stone is thrown vertically upwards with speed 14.7 m s⁻¹ from ground level. Taking up as positive, find the maximum height and the total time before it returns to the ground.

一块石头以 14.7 m s⁻¹ 的初速度从地面竖直上抛。取向上为正，求最大高度和它落回地面所需的总时间。

At the highest point v = 0. Use v² = u² + 2as:

最高点 v = 0。使用 v² = u² + 2as：

0² = 14.7² + 2(−9.8)s

Solving gives s = 14.7² ÷ 19.6 = 11.025 m, so the maximum height is approximately 11.0 m.

解得 s = 14.7² ÷ 19.6 = 11.025 m，因此最大高度约为 11.0 m。

For the total time, the displacement is s = 0 when the stone returns to the ground. Use s = ut +

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