位移wèiyí

Core Concept

Displacement (位移) is the straight-line change in position: it points from where you started to where you finished, and nothing about the route in between matters.

Its partner concept is distance (路程) — the total length of the actual path travelled. 路程和位移的计算 is a high-frequency CSCA knowledge point precisely because students conflate the two.

Definition

$\Delta x = x_{\text{final}} - x_{\text{initial}}$

• Vector (矢量): has magnitude and direction
• SI unit: metre ($\mathrm{m}$)
• Can be positive, negative, or zero

Displacement vs Distance

	Displacement 位移	Distance 路程
Type	vector	scalar
Depends on path?	no	yes
Can be zero while moving?	yes	no
Relationship	$\lvert\Delta x\rvert \leq$ distance	always $\geq \lvert\Delta x\rvert$

The two are equal only when the motion is in a straight line without reversing direction. Every other case has distance strictly greater.

Two Dimensions

Displacement adds like any vector. Perpendicular legs combine by the Pythagorean theorem:

$\lvert\Delta \vec{r}\rvert = \sqrt{(\Delta x)^2 + (\Delta y)^2}$

Walking $3\ \mathrm{m}$ east then $4\ \mathrm{m}$ north: distance $= 7\ \mathrm{m}$, displacement $= \sqrt{9+16} = 5\ \mathrm{m}$ north-east.

Reading It Off a Graph

On a position–time graph, displacement over an interval is the change in height between the endpoints.

On a velocity–time graph, displacement is the signed area under the curve — area below the axis counts as negative. Distance, by contrast, is the total unsigned area. This is the single most useful graph fact in CSCA kinematics.

CSCA Practice Problems

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

Example 1: Basic (Difficulty ★★☆☆☆)

An athlete runs one and a half laps of a circular track of circumference $400\ \mathrm{m}$. The distance and the magnitude of the displacement are:

Options:

• A. $600\ \mathrm{m}$ and $0$
• B. $600\ \mathrm{m}$ and about $127\ \mathrm{m}$
• C. $400\ \mathrm{m}$ and $200\ \mathrm{m}$
• D. $600\ \mathrm{m}$ and $200\ \mathrm{m}$

Detailed Solution:

1. Distance is the path length: $1.5 \times 400 = 600\ \mathrm{m}$.

2. One and a half laps ends diametrically opposite the start, so the displacement is one diameter.

3. From $C = 2\pi r = 400$: $r = \dfrac{400}{2\pi} \approx 63.7\ \mathrm{m}$, so the diameter is about $127\ \mathrm{m}$.

Option D is the trap for treating half a lap ($200\ \mathrm{m}$ of arc) as the displacement — but displacement is the straight line, not the arc.

Answer: B

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

A body moves at $+4\ \mathrm{m/s}$ for $3\ \mathrm{s}$, then at $-2\ \mathrm{m/s}$ for $4\ \mathrm{s}$. Find its displacement and the distance travelled.

Options:

• A. displacement $4\ \mathrm{m}$, distance $20\ \mathrm{m}$
• B. displacement $20\ \mathrm{m}$, distance $20\ \mathrm{m}$
• C. displacement $4\ \mathrm{m}$, distance $4\ \mathrm{m}$
• D. displacement $-4\ \mathrm{m}$, distance $20\ \mathrm{m}$

Detailed Solution:

1. First leg: $x_1 = (+4)(3) = +12\ \mathrm{m}$

2. Second leg: $x_2 = (-2)(4) = -8\ \mathrm{m}$

3. Displacement adds the signed contributions: $\Delta x = 12 + (-8) = +4\ \mathrm{m}$

4. Distance adds the magnitudes: $12 + 8 = 20\ \mathrm{m}$

On a velocity–time graph this is exactly signed area versus total area.

Answer: A

Common Mistakes

❌ Mistake 1: Using distance where displacement belongs

$\bar{v} = \Delta x/\Delta t$ and $x = v_0t + \frac{1}{2}at^2$ both produce displacement. Feeding them distance breaks any problem involving a reversal.

❌ Mistake 2: Adding path segments that point in opposite directions

Two $5\ \mathrm{m}$ moves in opposite directions give displacement $0$, not $10\ \mathrm{m}$.

❌ Mistake 3: Taking the arc as the displacement on a circular track

Displacement is always the straight chord between endpoints.

❌ Mistake 4: Ignoring sign when reading a velocity–time graph

Area below the time axis is negative displacement. Summing all areas as positive gives distance, not displacement.

Study Tips

1. ✅ Underline "distance" or "displacement" in the question before computing anything
2. ✅ Sketch the path — one drawing settles most of these questions
3. ✅ Remember $\lvert\Delta x\rvert \leq$ distance, with equality only for straight-line, no-reversal motion
4. ✅ On a $v$–$t$ graph: signed area = displacement, total area = distance
5. ✅ Combine perpendicular legs with Pythagoras, not by adding

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💡 Exam Tip: 路程和位移的计算 is among the top CSCA physics knowledge points. Questions almost always include a reversal or a closed loop — that is the whole test.