📚 Calculating Acceleration | 加速度的计算
Acceleration is one of the most frequently examined quantities in CIE A-Level Physics. It links force, motion and energy, and it appears in kinematics, dynamics and experimental analysis. A reliable method for calculating acceleration is therefore essential.
加速度是 CIE A-Level 物理中最常考查的物理量之一。它把力、运动和能量联系起来，出现在运动学、动力学和实验分析中。因此，掌握可靠的加速度计算方法至关重要。
1. Defining Acceleration | 加速度的定义
Acceleration is defined as the rate of change of velocity with respect to time. Since velocity is a vector, acceleration is also a vector; it has both magnitude and direction. The average acceleration a over a time interval Δt is given by:
加速度定义为速度随时间的变化率。由于速度是矢量，加速度也是矢量；它既有大小又有方向。在时间间隔 Δt 内的平均加速度 a 由下式给出：
a = Δv ÷ Δt = (v − u) ÷ t
where u is the initial velocity, v is the final velocity and t is the time taken. The SI unit of acceleration is metre per second squared, m s⁻².
其中 u 是初速度，v 是末速度，t 是所用时间。加速度的 SI 单位是米每二次方秒，即 m s⁻²。
An acceleration of 2 m s⁻² means the velocity increases by 2 m s⁻¹ every second. Always include the unit when quoting an acceleration value.
加速度为 2 m s⁻² 表示速度每秒增加 2 m s⁻¹。写出加速度数值时务必带上单位。
2. Average Acceleration vs Instantaneous Acceleration | 平均加速度与瞬时加速度
Average acceleration uses the total change in velocity over a finite time interval. Instantaneous acceleration is the acceleration at a specific instant, found by making Δt very small.
平均加速度使用有限时间间隔内速度的总变化量。瞬时加速度是某一特定时刻的加速度，通过让 Δt 趋于极小求得。
On a velocity-time graph, average acceleration is the slope of a chord between two points, while instantaneous acceleration is the slope of the tangent at one point.
在速度-时间图中，平均加速度是两点之间割线的斜率，而瞬时加速度是某一点处切线的斜率。
In many CIE calculations you may assume constant acceleration, so the average and instantaneous values are equal. This assumption is stated as ‘uniform acceleration’ in exam questions.
在许多 CIE 计算中，你可以假设加速度恒定，因此平均加速度与瞬时加速度相等。这一假设在考题中通常表述为“匀加速度”。
3. Sign and Direction | 正负号与方向
In one-dimensional motion, choose a positive direction and stick to it throughout the calculation. If u = 10 m s⁻¹ to the right and v = 4 m s⁻¹ to the left, taking right as positive gives u = +10 and v = −4, so a = (−4 − 10) ÷ t.
在一维运动中，选择一个正方向并始终坚持。若向右的初速度为 10 m s⁻¹，向左的末速度为 4 m s⁻¹，取向右为正，则 u = +10，v = −4，因此 a = (−4 − 10) ÷ t。
Acceleration can be negative even when an object is speeding up, if the object moves in the negative direction and its velocity becomes more negative. Use the signs of velocity and acceleration together: the same sign means speeding up, opposite signs mean slowing down.
当物体沿负方向运动且速度变得更负时，即使它在加速，加速度也可以为负。要结合速度和加速度的符号判断：符号相同表示加速，符号相反表示减速。
State your chosen positive direction clearly in CIE written answers. This avoids sign errors when velocities change direction.
在 CIE 书面作答中要清楚地说明所选正方向。这样可以避免速度方向改变时出现正负号错误。
4. Rearranging the Definition Formula | 定义公式的变形
The basic equation can be rearranged to find the final velocity: v = u + at. This applies only when acceleration is constant.
基本方程可以变形为求末速度：v = u + at。这仅在加速度恒定时适用。
v = u + at
For example, a cyclist starts at 3 m s⁻¹ and accelerates at 0.5 m s⁻² for 8 s: v = 3 + 0.5 × 8 = 7 m s⁻¹.
例如，一名自行车手以 3 m s⁻¹ 起步，以 0.5 m s⁻² 加速 8 秒：v = 3 + 0.5 × 8 = 7 m s⁻¹。
Always check units before substituting. If time is given in milliseconds or velocity in km h⁻¹, convert to SI units: seconds and metres per second.
代入前务必检查单位。如果时间以毫秒给出或速度以 km h⁻¹ 给出，要先转换为 SI 单位：秒和米每秒。
5. Velocity-Time Graphs | 速度-时间图
The gradient of a velocity-time graph gives acceleration. For a straight line, choose two widely spaced points on the line and use gradient = (v₂ − v₁) ÷ (t₂ − t₁). Do not use data points that lie off the best-fit line.
速度-时间图的斜率表示加速度。对于直线，选取直线上两个间距较大的点，使用斜率 = (v₂ − v₁) ÷ (t₂ − t₁)。不要使用偏离最佳拟合线的数据点。
The area under a velocity-time graph gives displacement, but this is not the same as acceleration. Keep the two skills separate: gradient for acceleration, area for displacement.
速度-时间图线下的面积表示位移，但这与加速度不同。要把两种技能区分开：斜率求加速度，面积求位移。
If the graph is curved, draw a tangent at the required time and find its gradient for instantaneous acceleration. A changing gradient means the acceleration is changing.
如果图线是曲线，在所需时刻作切线，并求其斜率以获得瞬时加速度。斜率变化意味着加速度在变化。
6. Equations of Motion (SUVAT) | 运动学方程（SUVAT）
For constant acceleration in a straight line, the four equations of motion are:
对于直线匀加速运动，四个运动学方程为：
v = u + at s = (u + v)t ÷ 2 s = ut + ½at² v² = u² + 2as
where s is displacement, u is initial velocity, v is final velocity, a is acceleration and t is time.
其中 s 是位移，u 是初速度，v 是末速度，a 是加速度，t 是时间。
| Symbol | Quantity | SI unit |
| s | displacement / 位移 | m |
| u | initial velocity / 初速度 | m s⁻¹ |
| v | final velocity / 末速度 | m s⁻¹ |
| a | acceleration / 加速度 | m s⁻² |
| t | time / 时间 | s |
To choose the right equation, identify which quantity is not given and not required. If a question gives u, v and t and asks for a, use a = (v − u) ÷ t. If it gives u, a and t and asks for s, use s = ut + ½at².
要选择正确的方程，先确定哪个量未给出且不需要。如果题目给出 u、v 和 t 并要求 a，使用 a = (v − u) ÷ t。如果给出 u、a 和 t 并要求 s，使用 s = ut + ½at²。
7. Worked Example: Braking Car | 计算实例：刹车的汽车
A car travelling at 24 m s⁻¹ brakes uniformly and stops in 6.0 s. Calculate the acceleration and the braking distance.
一辆以 24 m s⁻¹ 行驶的汽车均匀刹车，在 6.0 秒内停下。计算加速度和刹车距离。
Take the original direction as positive, so u = +24 m s⁻¹, v = 0 and t = 6.0 s. Acceleration: a = (v − u) ÷ t = (0 − 24) ÷ 6.0 = −4.0 m s⁻². The minus sign shows the acceleration is opposite to the motion.
取原运动方向为正，则 u = +24 m s⁻¹，v = 0，t = 6.0 s。加速度：a = (v − u) ÷ t = (0 − 24) ÷ 6.0 = −4.0 m s⁻²。负号表示加速度与运动方向相反。
Braking distance: s = (u + v)t ÷ 2 = (24 + 0) × 6.0 ÷ 2 = 72 m.
刹车距离：s = (u + v)t ÷ 2 = (24 + 0) × 6.0 ÷ 2 = 72 m。
Notice that the negative acceleration does not make the distance negative. Displacement is positive because the car continues moving in the positive direction while slowing down.
注意负加速度不会使距离变为负值。位移为正，因为汽车在减速过程中仍沿正方向运动。
8. Free Fall and g | 自由落体与重力加速度 g
Near Earth’s surface, all objects in free fall have the same downward acceleration g = 9.81 m s⁻², provided air resistance is negligible. If you take downward as positive, a = +9.81 m s⁻²; if upward is positive, a = −9.81 m s⁻².
在地球表面附近，若空气阻力可忽略，所有自由落体都具有相同的向下加速度 g = 9.81 m s⁻²。如果取向下为正，a = +9.81 m s⁻²；如果取向上为正，a = −9.81 m s⁻²。
A stone dropped from rest falls for 3.0 s. Using s =
Published by TutorHao | A-Level Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导