速度sùdù

Core Concept

Velocity (速度) is the rate at which position changes. In Chinese physics texts the word covers two distinct English ideas, and CSCA questions exploit the difference constantly:

• Speed (速率) — a scalar: how fast, no direction
• Velocity (速度) — a vector: how fast and in which direction

Average Velocity

$\bar{v} = \frac{\Delta x}{\Delta t}$

Note that this uses displacement $\Delta x$, not distance travelled. That single distinction generates most of the exam traps on this topic.

Average Speed

$\text{average speed} = \frac{\text{total distance}}{\Delta t}$

A runner completing one lap of a $400\ \mathrm{m}$ track in $50\ \mathrm{s}$ has:

• Average speed = $400/50 = 8\ \mathrm{m/s}$
• Average velocity = $0/50 = 0$, because they finished where they started

Instantaneous Velocity

The velocity at one instant — the limit of $\Delta x/\Delta t$ as $\Delta t \to 0$. This is what a speedometer reads, and what "at $t = 3\ \mathrm{s}$ the velocity is…" refers to.

For uniform acceleration, instantaneous velocity follows:

$v = v_0 + at$

A Useful Shortcut

Under uniform acceleration only, the average velocity equals the mean of the endpoints:

$\bar{v} = \frac{v_0 + v}{2}$

This turns many two-step problems into one line. It is not valid when acceleration varies.

CSCA Practice Problems

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

Example 1: Basic (Difficulty ★★☆☆☆)

A car travels $60\ \mathrm{km}$ east in $1\ \mathrm{h}$, then $60\ \mathrm{km}$ back west in $1\ \mathrm{h}$. Over the whole trip:

Options:

• A. average speed $60\ \mathrm{km/h}$, average velocity $60\ \mathrm{km/h}$
• B. average speed $60\ \mathrm{km/h}$, average velocity $0$
• C. average speed $0$, average velocity $0$
• D. average speed $120\ \mathrm{km/h}$, average velocity $0$

Detailed Solution:

1. Total distance = $60 + 60 = 120\ \mathrm{km}$; total time = $2\ \mathrm{h}$. $\text{average speed} = \frac{120}{2} = 60\ \mathrm{km/h}$

2. The car ends where it began, so displacement $= 0$: $\bar{v} = \frac{0}{2} = 0$

Answer: B

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

A body starts from rest and accelerates uniformly, reaching $12\ \mathrm{m/s}$ after travelling $36\ \mathrm{m}$. How long did it take?

Options:

• A. $3\ \mathrm{s}$
• B. $6\ \mathrm{s}$
• C. $9\ \mathrm{s}$
• D. $12\ \mathrm{s}$

Detailed Solution:

Use the average-velocity shortcut instead of solving for $a$ first:

1. Uniform acceleration from rest, so $\bar{v} = \frac{v_0 + v}{2} = \frac{0 + 12}{2} = 6\ \mathrm{m/s}$

2. Then $t = \frac{\Delta x}{\bar{v}} = \frac{36}{6} = 6\ \mathrm{s}$

Answer: B

Check the long way: $v^2 = 2ax \Rightarrow 144 = 2a(36) \Rightarrow a = 2\ \mathrm{m/s^2}$, then $t = v/a = 12/2 = 6\ \mathrm{s}$ ✓ — same answer, twice the work.

Common Mistakes

❌ Mistake 1: Using distance in the average-velocity formula

$\bar{v}$ uses displacement. Any closed loop gives zero average velocity no matter how far the object actually travelled.

❌ Mistake 2: Averaging two speeds directly

For equal distances at $v_1$ and $v_2$, the average speed is not $\frac{v_1+v_2}{2}$ — it is the harmonic mean $\frac{2v_1v_2}{v_1+v_2}$, because more time is spent at the slower speed. Averaging endpoints is valid only for uniform acceleration over equal time.

❌ Mistake 3: Treating velocity as always positive

Velocity is a vector; a negative value simply means motion in the negative direction.

Study Tips

1. ✅ Ask "distance or displacement?" before every average calculation
2. ✅ Memorise $\bar{v} = \frac{v_0+v}{2}$ and the condition it needs (uniform acceleration)
3. ✅ Watch for closed paths — round trips are a giveaway that average velocity is $0$
4. ✅ Keep speed and velocity separate even though Chinese uses 速率 and 速度 closely
5. ✅ Label your positive direction and let negative values do their job

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💡 Exam Tip: 平均速度 and 瞬时速度 are both recurring CSCA knowledge points, and 路程和位移的计算 sits right beside them — most questions are really testing whether you noticed the distance-versus-displacement switch.