📚 Methods of Calculating Acceleration | 加速度的计算方法
Acceleration is one of the most frequently tested concepts in CIE A-Level Physics. You need to be able to calculate it from data, velocity-time graphs, experimental measurements, equations of motion, and Newton's second law. This article presents each method clearly, with worked examples and exam-focused advice.
加速度是CIE A-Level物理中最常考的概念之一。你需要能够从数据、速度-时间图像、实验测量、运动学方程以及牛顿第二定律中计算加速度。本文将逐一介绍这些方法，并给出例题和考试建议。
1. Defining Acceleration | 加速度的定义
Acceleration is defined as the rate of change of velocity with respect to time. Because velocity is a vector quantity, acceleration is also a vector quantity; it has both magnitude and direction.
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，既有大小也有方向。
For motion in a straight line with constant acceleration:
对于直线匀加速运动：
a = Δv / Δt = (v − u) / t
Here, u is the initial velocity, v is the final velocity, and t is the time taken. The SI unit of acceleration is metres per second squared, written as m/s².
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，写作m/s²。
tive, the object is accelerating in the direction opposite to the chosen positive direction.
一个常见的等价形式是使用速度变化量Δv = v − u。如果Δv为负，则物体在选定的正方向的反方向上加速。
2. Uniform Acceleration Equations | 匀加速运动方程
When acceleration is constant, the equations of motion known as the suvat equations can be applied. They link displacement s, initial velocity u, final velocity v, acceleration a, and time t.
当加速度恒定时，可以使用称为suvat方程组的运动学公式。它们联系了位移s、初速度u、末速度v、加速度a和时间t。
The most useful forms are:
最常用的形式如下：
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t
to find. For example, if you know u, v and s but not t, use v² = u² + 2as and solve for a.
选择包含已知量和待求量的方程。例如，若已知u、v和s但未知t，就使用v² = u² + 2as并解出a。
These equations are valid only while acceleration is uniform. If acceleration changes with time, suvat equations cannot be used directly.
这些方程仅在加速度恒定不变时成立。如果加速度随时间变化，则不能直接使用suvat方程。
3. Velocity-Time Graphs | 速度-时间图像
The gradient of a velocity-time graph represents acceleration. For a straight line, the acceleration is constant and can be calculated as:
速度-时间图像的斜率表示加速度。对于直线，加速度恒定，计算公式为：
a = (v₂ − v₁) / (t₂ − t₁)
For a curved graph, the acceleration is changing. In that case, draw a tangent to the curve at the point of interest; the gradient of this tangent gives the instantaneous acceleration.
对于曲线图像，加速度是变化的。此时应在目标位置作曲线的切线，切线的斜率就是瞬时加速度。
| Shape of graph | Meaning |
| Straight line sloping upward | Constant positive acceleration |
| Straight line sloping downward | Constant negative acceleration, i.e. deceleration in the positive direction |
| Horizontal line | Zero acceleration, constant velocity |
| Curved line | Changing acceleration; gradient changes continuously |
Do not confuse the gradient with the area under the graph. The area under a velocity-time graph represents displacement, not acceleration.
不要将斜率与图像下方面积混淆。速度-时间图像下方面积表示位移，而不是加速度。
4. Light Gates and Ticker Timers | 光电门与打点计时器
In the laboratory, acceleration is often measured using two light gates and a data logger. A card of known length L is passed through each gate. The speed at each gate is found from:
在实验室中，常使用两个光电门和数据记录器测量加速度。已知长度为L的挡光片经过每个光电门。每个光电门处的速度由下式得到：
v = L / t
If the two light gates are separated by distance s, and the speeds at the gates are u and v, then the acceleration between them is:
如果两个光电门相距s，通过两门时的速度分别为u和v，则它们之间的加速度为：
a = (v² − u²) / (2s)
vely, if the time between the two gates is measured directly, you can use a = (v − u) / t.
该方程由v² = u² + 2as变形得到。另外，如果直接测量通过两光电门之间的时间，也可使用a = (v − u) / t。
A ticker timer can also measure acceleration. It prints dots at a constant frequency, often 50 Hz, so each dot interval is 0.02 s. Measure the distance between two dots to find speed:
打点计时器也可用来测量加速度。它以恒定频率（通常为50 Hz）在纸带上打点，因此相邻两点间的时间间隔为0.02 s。测量两点间距离可得到速度：
v = distance between dots / time interval
nd v₂, and the time between the midpoints of those intervals is t, then a = (v₂ − v₁) / t.
然后测量连续两个时间间隔的速度。如果速度分别为v₁和v₂，两个间隔中点之间的时间为t，则a = (v₂ − v₁) / t。
5. Acceleration on an Inclined Plane | 斜面上的加速度
An object on a smooth inclined plane experiences acceleration due to the component of its weight acting parallel to the plane. If the plane is at angle θ to the horizontal, this component is mg sinθ.
光滑斜面上的物体受到重力沿斜面方向的分量作用而产生加速度。若斜面与水平方向夹角为θ，则该分量为mg sinθ。
Using Newton's second law:
由牛顿第二定律：
a = F / m = mg sinθ / m = g sinθ
So on a frictionless incline, the acceleration depends only on g and the angle θ. It does not depend on the mass of the object.
因此，在无摩擦斜面上，加速度只取决于g和角度θ，与物体质量无关。
If friction of magnitude f acts up the slope, the net force becomes mg sinθ − f, so the acceleration is:
如果存在大小为f、方向沿斜面向上的摩擦力，则合力为mg sinθ − f，因此加速度为：
a = g sinθ − f / m
This explains why a rough incline produces a smaller acceleration than the ideal value g sinθ.
这解释了为什么粗糙斜面上的加速度会小于理想值g sinθ。
6. Free Fall and Gravitational Acceleration | 自由落体与重力加速度
eleration is the gravitational acceleration g. On Earth, g ≈ 9.81 m/s², directed downward.
当物体在空气阻力可忽略的情况下自由下落时，其加速度为重力加速度g。在地球表面，g ≈ 9.81 m/s²，方向向下。
For an object released from rest, the equations of motion become:
对于从静止释放的物体，运动方程变为：
v = g t
s = ½ g t²
v² = 2 g s
Here s is the vertical distance fallen. These equations are often used in projectile motion problems where vertical motion is analysed separately from horizontal motion.
其中s是下落的竖直距离。这些方程常用于抛体运动中，此时竖直方向与水平方向的运动分别分析。
If air resistance becomes significant, the resultant downward force decreases, so the acceleration decreases. Eventually the object reaches terminal velocity, where acceleration becomes zero.
当空气阻力不可忽略时，向下的合力减小，因此加速度减小。最终物体达到终端速度，此时加速度变为零。
7. Newton's Second Law | 牛顿第二定律
Newton's second law states that the net force acting on an object is equal to the product of its mass and acceleration:
牛顿第二定律指出，物体所受合力等于其质量与加速度的乘积：
Fₙₑₜ = m a
Rearranging gives:
移项得：
a = Fₙₑₜ / m
It is important to use the resultant force, not the applied force alone. If forces act in opposite directions, subtract them before calculating acceleration.
注意必须使用合力，而不能只使用施加的某一个力。如果力方向相反，要先相减再计算加速度。
For example, if a resultant force of 12 N acts on a 4.0 kg mass, the acceleration is:
例如，若12 N的合力作用在4.0 kg的物体上，加速度为：
a = 12 N / 4.0 kg = 3.0 m/s²
This method is especially useful when combined with weight, friction, tension or normal contact forces in mechanics problems.
这种方法在力学问题中特别有用，可结合重力、摩擦力、张力和法向接触力一起分析。
8. Average Acceleration vs Instantaneous Acceleration | 平均加速度与瞬时加速度
Average acceleration is calculated over a finite time interval using the total change in velocity divided by the total time:
平均加速度是在一段有限时间内，用总速度变化量除以总时间计算：
aₐᵥ = (v − u) / t
Instantaneous acceleration is the acceleration at a particular instant. It can be found by making the time interval very small, or by drawing a tangent to a velocity-time graph and finding its gradient.
瞬时加速度是某一时刻的加速度。可以通过使时间间隔非常小得到，也可以在速度-时间图像上作曲线的切线并求其斜率。
These two values are equal only when acceleration is constant. When acceleration changes, the average and instantaneous values are different.
只有加速度恒定时，平均加速度才等于瞬时加速度。当加速度变化时，两者是不同的。
In CIE exam questions, read carefully whether the question asks for average acceleration or instantaneous acceleration. Using the wrong value is a common reason for losing marks.
在CIE考试中，要仔细看题目要求的是平均加速度还是瞬时加速度。用错数值是常见的失分原因。
9. Sign Convention and Direction | 正负号与方向
Because acceleration is a vector, you must choose a positive direction before calculating it. Once the direction is chosen, all velocities and accelerations must be signed consistently.
因为加速度是矢量，计算前必须先规定正方向。一旦确定方向，所有速度和加速度都必须带符号并保持一致。
ball slows down as it moves upward because velocity and acceleration have opposite signs.
例如，规定向上为正。竖直上抛的小球初速度u > 0，重力加速度为a = −9.81 m/s²。小球上升时速度与加速度方向相反，因此减速。
At the highest point, the velocity is zero, but the acceleration is still −9.81 m/s². A common mistake is to say that acceleration is zero at the top of the flight; this is incorrect.
在最高点，速度为零，但加速度仍为−9.81 m/s²。一个常见错误是认为物体在最高点加速度为零，这是不对的。
When an object is slowing down, the acceleration is opposite to the velocity. In many problems, this will simply appear as a negative value, depending on your sign convention.
当物体减速时，加速度方向与速度方向相反。在很多题目中，根据所选正方向，这会表现为负值。
10. Worked Examples | 例题
Example 1: Uniform acceleration
例1：匀加速运动
A car increases its velocity from 5 m/s to 25 m/s in 4.0 s. Calculate its average acceleration.
一辆汽车在4.0 s内速度从5 m/s增加到25 m/s。求它的平均加速度。
a = (v − u) / t = (25 − 5) / 4.0 = 5.0 m/s²
The acceleration is 5.0 m/s² in the direction of motion.
加速度大小为5.0 m/s²，方向与运动方向相同。
Example 2: Using two light gates
例2：使用两个光电门
A trolley passes the first gate at 0.20 m/s and the second gate at 0.80 m/s. The gates are separated by 0.50 m. Find the acceleration.
小车通过第一个光电门时速度为0.20 m/s，通过第二个光电门时速度为0.80 m/s。两光电门相距0.50 m。求加速度。
v² = u² + 2as
a = (v² − u²) / (2s) = (0.80² − 0.20²) / (2 × 0.50)
a = (0.64 − 0.04) / 1.0 = 0.60 m/s²
Example 3: Newton's second law
例3：牛顿第二定律
A resultant force of 18 N acts on a mass of 6.0 kg. What is the acceleration?
一个大小为18 N的合力作用在6.0 kg的物体上。加速度是多少？
a = Fₙₑₜ / m = 18 / 6.0 = 3.0 m/s²
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
-
Do not confuse velocity with acceleration. Velocity is how fast an object moves; acceleration is how quickly the velocity changes.
不要混淆速度与加速度。速度是物体运动得快慢，加速度是速度变化的快慢。
-
Always convert units before substituting into equations. If speed is given in km/h, divide by 3.6 to change to m/s.
代入公式前一定要换算单位。若速度以km/h给出，除以3.6即可换算为m/s。
-
Use the suvat equations only for uniform acceleration. For non-uniform acceleration, use velocity-time graphs or average acceleration over a short interval.
suvat方程只适用于匀加速运动。对于非匀加速运动，应使用速度-时间图像或短时间内的平均加速度。
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