Motion graphs and acceleration
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| velocity/vəˈlɒsɪti/ | 速度 | sù dù |
| acceleration/əkˌseləˈreɪʃn/ | 加速度 | jiā sù dù |
What would explain this observation?
- A speedometer gives a reading at one moment. A journey average can hide stops and rapid changes in speed.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Displacement includes direction; distance counts total path length. Velocity 速度 is change in displacement per time. Acceleration 加速度 is change in velocity per time.
- velocity: Rate of change of displacement; acceleration: Rate of change of velocity.
What does area under a velocity-time graph represent?
The gradient of a displacement-time graph is velocity. The area under a velocity-time graph gives displacement. A constant-acceleration formula is valid only when its assumption is justified.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- The gradient of a displacement-time graph is velocity. The area under a velocity-time graph gives displacement. A constant-acceleration formula is valid only when its assumption is justified.
- Choose a positive direction and state it. Use a light gate or video with a known scale and frame interval for repeatable motion measurements. Avoid assuming hand timing is exact over very short intervals.
Which two habits make the investigation or model in this case more defensible?
Choose a positive direction and state it. Use a light gate or video with a known scale and frame interval for repeatable motion measurements. Avoid assuming hand timing is exact over very short intervals.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: velocity rises uniformly from 2 to 10 metres per second in 4 s. a = (v-u)/t. a = (10-2)/4 = 2 metres per second squared. Displacement is trapezium area: s = (u+v)t/2 = (2+10)×4/2 = 24 m.
Velocity changes from 3 to 15 metres per second in 6 s. Find acceleration. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
Velocity changes from 3 to 15 metres per second in 6 s. Find acceleration.
The result is 2 m/s². Known: velocity rises uniformly from 2 to 10 metres per second in 4 s. a = (v-u)/t. a = (10-2)/4 = 2 metres per second squared. Displacement is trapezium area: s = (u+v)t/2 = (2+10)×4/2 = 24 m.
Check the conclusion and its limits
- Negative velocity indicates direction under the chosen sign convention; it does not necessarily mean slowing down.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
A negative velocity always means the object is slowing down. This claim is false: Negative velocity indicates direction under the chosen sign convention; it does not necessarily mean slowing down.
Motion graphs and acceleration: The gradient of a displacement-time graph is velocity. The area under a velocity-time graph gives displacement. A constant-acceleration formula is valid only when its assumption is justified.
A negative velocity always means the object is slowing down.
Negative velocity indicates direction under the chosen sign convention; it does not necessarily mean slowing down.
Rate of change of displacement: write the technical term.
velocity means Rate of change of displacement.