Displacement, Velocity, and Acceleration
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| distance/ˈdɪstəns/ | quãng đường |
| position/pəˈzɪʃn/ | vị trí |
| displacement/dɪˈspleɪsmənt/ | tích dịch chuyển |
| scalar/ˈskeɪlə/ | đại lượng vô hướng |
| vector/ˈvektə/ | đại lượng có hướng |
| velocity/vəˈlɒsɪti/ | vận tốc |
| average velocity/ˈævrɪdʒ vəˈlɒsɪti/ | tốc độ trung bình |
| speed/spiːd/ | tốc độ |
| magnitude/ˈmæɡnɪtjuːd/ | độ lớn |
| instantaneous velocity/ˌɪnstənˈteɪnɪəs vəˈlɒsɪti/ | tốc độ tức thời |
| acceleration/əkˌseləˈreɪʃn/ | gia tốc |
One full lap — so how far from the start?
- A runner sprints one lap of a 400 m track and stops exactly where they began.
- Their distance 距离 travelled is $400\ \text{m}$. Their change in position is zero.
- Two honest answers to "how far?" — because one counts the path, the other counts the change in place.
- Kinematics needs both, and never mixes them up.
Distance vs displacement
- Distance (a scalar 标量) is the total path length — always positive, it only grows.
- Displacement 位移 (a vector 矢量) is the straight-line change in position 位置.
- We write it $\Delta x = x_f - x_i$ — final position minus initial position.
- On the lap, distance $= 400\ \text{m}$ but displacement $\Delta x = 0$.
A runner completes exactly one lap of a 400 m track. What is their displacement?
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Displacement is the change in position. The runner ends where they started, so $\Delta x = 0$. The distance travelled is $400\ \text{m}$.
Velocity: how fast position changes
- Velocity 速度 is displacement per unit time — it carries a direction.
- Average velocity 平均速度: $\displaystyle v_{\text{avg}} = \frac{\Delta x}{\Delta t}$.
- Speed 速率 is the magnitude 大小 of velocity — a scalar, no direction.
- Drive 300 km in 5 h with detours: high average speed, but a small average velocity if you end up near home.
A cyclist has a displacement of $240\ \text{m}$ in $30\ \text{s}$. What is the average velocity, in $\tfrac{\text{m}}{\text{s}}$?
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$v_{\text{avg}} = \dfrac{\Delta x}{\Delta t} = \dfrac{240}{30} = 8\ \tfrac{\text{m}}{\text{s}}$.
Instantaneous velocity
- Average velocity smears over a whole interval; instantaneous velocity 瞬时速度 is the value at one instant.
- It is what $\dfrac{\Delta x}{\Delta t}$ approaches as $\Delta t$ shrinks toward zero.
- On a position–time graph, it is the slope of the tangent at that point.
- A speedometer reads instantaneous speed — the size of the instantaneous velocity.
Acceleration: how fast velocity changes
- Acceleration 加速度 is the rate of change of velocity: $\displaystyle a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{\Delta t}$.
- It is a vector, measured in $\tfrac{\text{m}}{\text{s}^2}$.
- Speeding up, slowing down, and turning are all accelerations — each changes the velocity vector.
- The units stack neatly: position $\text{m}$, velocity $\tfrac{\text{m}}{\text{s}}$, acceleration $\tfrac{\text{m}}{\text{s}^2}$.
Build the motion from u and a
Change the starting velocity and acceleration, and watch how the position and velocity graphs respond.
Acceleration is the rate of change of ____.
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By definition $a = \Delta v / \Delta t$ — the change in velocity per unit time.
Select all the quantities that are vectors.
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Displacement, velocity and acceleration all carry a direction. Distance and speed are scalars (magnitude only).
A negative acceleration does not always mean "slowing down". An object slows only when its acceleration points opposite to its velocity. If both are negative (e.g. falling faster downward), it is speeding up.
A negative acceleration always means the object is slowing down.
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An object slows only when acceleration points opposite to velocity. A negative acceleration on a negative (downward) velocity speeds the object up.
A car speeds up from $8\ \tfrac{\text{m}}{\text{s}}$ to $20\ \tfrac{\text{m}}{\text{s}}$ in $4\ \text{s}$.
- $\Delta v = 20 - 8 = 12\ \tfrac{\text{m}}{\text{s}}$.
- $a = \dfrac{\Delta v}{\Delta t} = \dfrac{12}{4} = 3\ \tfrac{\text{m}}{\text{s}^2}$.
The acceleration points the same way as the velocity, so the car speeds up.
A car speeds up from $8\ \tfrac{\text{m}}{\text{s}}$ to $20\ \tfrac{\text{m}}{\text{s}}$ in $4\ \text{s}$. What is its acceleration, in $\tfrac{\text{m}}{\text{s}^2}$?
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$a = \dfrac{v_f - v_i}{\Delta t} = \dfrac{20 - 8}{4} = 3\ \tfrac{\text{m}}{\text{s}^2}$.
Distance (scalar) counts the path; displacement $\Delta x = x_f - x_i$ (vector) counts the change in position. Velocity $= \Delta x / \Delta t$; speed is its magnitude. Acceleration $= \Delta v / \Delta t$ — how fast velocity changes, including turning.
Write from an acceleration of 4
- A cart speeds up from 6 m/s to 18 m/s in 3 s.
- Find the acceleration.
A cart speeds up from 6 m/s to 18 m/s in 3 s. What is its acceleration, in m/s squared?
12 divided by 3 is 4.
What is the acceleration?
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The speed rises by 12 m/s in 3 s. Other options — 3 m/s squared (the lesson car); 12 m/s squared (the speed change used as the acceleration).