加速度jiāsùdù

Core Concept

Acceleration (加速度) is the rate at which velocity changes. It is the single most frequently tested concept in CSCA physics kinematics — 加速度的计算 and 加速度的定义 together are the highest-frequency knowledge points in the physics question bank.

The crucial idea, and the one exam questions probe hardest: acceleration describes how velocity changes, not how fast something is moving. A car moving at a constant $100\ \mathrm{km/h}$ has zero acceleration. A car momentarily at rest at the top of a bounce still has acceleration $g$ downward.

Mathematical Definition

$a = \frac{\Delta v}{\Delta t} = \frac{v - v_0}{t}$

• $v_0$ — initial velocity (初速度)
• $v$ — final velocity (末速度)
• $t$ — elapsed time
• SI unit: $\mathrm{m/s^2}$ (metres per second squared)

Acceleration is a vector (矢量): it has both magnitude and direction. Its sign matters and is set by the direction you choose as positive.

Reading the Sign

$v$ and $a$	Motion
same sign	speeding up (加速)
opposite signs	slowing down (减速)
$a = 0$	constant velocity

Negative acceleration does not mean "slowing down". An object moving in the $-x$ direction and speeding up has negative velocity and negative acceleration. Whether it speeds up or slows down depends on whether the two signs agree.

The Kinematic Equations

For uniform acceleration (匀变速直线运动), three equations cover nearly every CSCA question:

$v = v_0 + at$

$x = v_0 t + \frac{1}{2}at^2$

$v^2 = v_0^2 + 2ax$

The third is the one students forget, and it is the one to reach for whenever a question gives you distances and speeds but no time.

Average vs Instantaneous

• Average acceleration (平均加速度): $\bar{a} = \dfrac{\Delta v}{\Delta t}$ over a finite interval
• Instantaneous acceleration (瞬时加速度): the limit as $\Delta t \to 0$

Under uniform acceleration the two are equal — which is why exam problems almost always specify 匀变速.

Free Fall

A body under gravity alone has constant acceleration:

$a = g \approx 9.8\ \mathrm{m/s^2} \quad (\text{often } 10\ \mathrm{m/s^2} \text{ in CSCA problems})$

directed downward, independent of the object's mass. Substituting $v_0 = 0$ into the equations above gives the free-fall results directly:

$v = gt, \qquad h = \frac{1}{2}gt^2, \qquad v^2 = 2gh$

CSCA Practice Problems

💡 Note: These problems follow the CSCA exam syllabus and Chinese standardized test formats.

Example 1: Basic (Difficulty ★★☆☆☆)

A car accelerates uniformly from $5\ \mathrm{m/s}$ to $25\ \mathrm{m/s}$ in $4\ \mathrm{s}$. Its acceleration is:

Options:

• A. $4\ \mathrm{m/s^2}$
• B. $5\ \mathrm{m/s^2}$
• C. $6.25\ \mathrm{m/s^2}$
• D. $7.5\ \mathrm{m/s^2}$

Detailed Solution:

Apply the definition directly:

$a = \frac{v - v_0}{t} = \frac{25 - 5}{4} = \frac{20}{4} = 5\ \mathrm{m/s^2}$

Option C is the trap for dividing the final velocity by the time ($25/4$) instead of the change in velocity — a slip that comes from reading "accelerates to $25\ \mathrm{m/s}$" as if it started from rest.

Answer: B

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Example 2: Intermediate (Difficulty ★★★☆☆)

A train braking uniformly at $2\ \mathrm{m/s^2}$ takes $100\ \mathrm{m}$ to stop. What was its initial speed?

Options:

• A. $10\ \mathrm{m/s}$
• B. $14.1\ \mathrm{m/s}$
• C. $20\ \mathrm{m/s}$
• D. $40\ \mathrm{m/s}$

Detailed Solution:

1. The question gives distance and acceleration but no time, which signals the third kinematic equation: $v^2 = v_0^2 + 2ax$

2. The train stops, so $v = 0$. Braking opposes the motion, so with forward as positive, $a = -2\ \mathrm{m/s^2}$: $0 = v_0^2 + 2(-2)(100)$

3. Solve: $v_0^2 = 400 \implies v_0 = 20\ \mathrm{m/s}$

4. Sanity check: at $20\ \mathrm{m/s}$ decelerating at $2\ \mathrm{m/s^2}$, stopping takes $10\ \mathrm{s}$, and average speed $10\ \mathrm{m/s} \times 10\ \mathrm{s} = 100\ \mathrm{m}$ ✓

Answer: C

Why not use $x = v_0t + \frac{1}{2}at^2$? You could, but you would first have to find $t$ from another equation — two steps where one suffices. Recognising "distance given, time absent" is worth real marks under time pressure.

Common Mistakes

❌ Mistake 1: Confusing acceleration with velocity

Large velocity does not imply large acceleration. A jet cruising at $250\ \mathrm{m/s}$ has $a = 0$; a car pulling away from a light at $5\ \mathrm{m/s}$ may have $a = 3\ \mathrm{m/s^2}$.

❌ Mistake 2: Assuming $a = 0$ when $v = 0$

At the peak of a vertical throw the velocity is momentarily zero, but the acceleration is still $g$ downward. This is a classic CSCA multiple-choice trap.

❌ Mistake 3: Sign errors on deceleration

Pick a positive direction, state it, and keep it for the whole problem. Braking while moving forward means $a < 0$.

❌ Mistake 4: Using uniform-acceleration equations when acceleration is not uniform

The three equations above are valid only for constant $a$. If a question describes changing acceleration, they do not apply.

Study Tips

1. ✅ Write $a = \Delta v / \Delta t$ first — the change in velocity, never the final velocity alone
2. ✅ Choose and label a positive direction before substituting anything
3. ✅ When time is missing, reach for $v^2 = v_0^2 + 2ax$
4. ✅ Remember $v = 0$ and $a = 0$ are independent — the throw-at-its-peak case proves it
5. ✅ Check units: $\mathrm{m/s^2}$, not $\mathrm{m/s}$

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💡 Exam Tip: 加速度的计算 and 加速度的定义 are the two highest-frequency knowledge points in CSCA physics. Most are one-step definition problems — the marks go to whoever keeps the signs straight.