Cinématique
Baccalauréat international · IB Diploma · Physics · HL Sujet 1 4:00 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
A speedometer gives a reading at one moment.
速度计显示的是某一时刻的瞬时读数。
A journey average can hide stops and rapid changes in speed.
而整个行程的平均值可能会掩盖中途停车和速度的剧烈变化。
Displacement includes direction; distance counts total path length.
位移包含方向；路程计算总路径长度。
Velocity is change in displacement per time.
速度是单位时间内的位移变化量。
Acceleration is change in velocity per time.
加速度是单位时间内的速度变化量。
Velocity is the rate of change of displacement.
速度是位移的变化率。
Acceleration is the rate of change of velocity.
加速度是速度的变化率。
The gradient of a displacement-time graph is velocity.
位移-时间图像的斜率表示速度。
The area under a velocity-time graph gives displacement.
速度-时间图像下的面积表示位移。
A constant-acceleration formula is valid only when its assumption is justified.
仅当前提条件成立时，匀加速运动公式才适用。
In the lab: Choose a positive direction and state it.
在实验室中，选择一个正方向并加以说明。
Use a light gate or video with a known scale and frame interval for repeatable motion measurements.
使用光电门或带有已知标尺和帧间隔的视频进行可重复的运动测量。
Avoid assuming hand timing is exact over very short intervals.
避免假设手动计时在极短的时间间隔内是精确的。
Known: velocity rises uniformly from 2 to 10 metres per second in 4 s.
已知：速度在 4 秒内从每秒 2 米均匀增加到每秒 10 米。
a equals v minus u, all divided by t.
a 等于 v 减 u，再除以 t。
a equals 10 minus 2, all divided by 4, which is 2 metres per second squared.
a 等于 10 减 2，再除以 4，即每二次方秒 2 米。
Displacement is trapezium area: s equals u plus v, times t, all over 2.
位移等于梯形面积：s 等于 u 加 v，乘以 t，再除以 2。
That is 2 plus 10, times 4, all over 2, which equals 24 m.
即 2 加 10，乘以 4，再除以 2，等于 24 米。
Your turn: Velocity changes from 3 to 15 metres per second in 6 s.
请作答。 速度从 3 米每秒变为 15 米每秒，用时 6 秒。
Find acceleration.
求加速度。
Pause the video and work it out.
请暂停视频，自己算一算。
Check: 15 minus 3 is 12; divided by 6 seconds, the acceleration is 2 metres per second squared.
核对：15 − 3 = 12，除以 6 秒，加速度为 2 米每二次方秒。
Negative velocity indicates direction under the chosen sign convention; it does not necessarily mean slowing down.
负速度表示所选符号约定下的方向，并不一定意味着减速。
Not true: A negative velocity always means the object is slowing down.
不正确。 负速度并不总是意味着物体在减速。
Negative velocity means motion opposite to the chosen positive direction.
负速度表示运动方向与所选正方向相反。
The object slows down only when the acceleration opposes the velocity; a negative velocity with negative acceleration is speeding up.
只有当加速度与速度方向相反时，物体才会减速。 负速度与负加速度同时存在时，物体是在加速。
Displacement includes direction; distance counts total path length.
位移包含方向；路程计算总路径长度。
The gradient of a displacement-time graph is velocity.
位移-时间图像的斜率即为速度。
Watch for negative velocity indicates direction under the chosen sign convention; it does not necessarily mean slowing down.
请注意：负速度表示所选符号约定下的方向，并不一定意味着减速。
Velocity changes from 4 to 22 metres per second in 3 s.
速度从 4 米每秒变化到 22 米每秒，用时 3 秒。
Find the acceleration and the displacement.
求加速度和位移。
A negative velocity is a direction.
负速度表示方向。
The result is an acceleration of 6 metres per second squared and a displacement of 39 metres.
结果是加速度为6米每二次方秒，位移为39米。
The velocity rises steadily from 2 to 10 metres per second in 4 seconds.
速度在4秒内从每秒2米均匀增加到每秒10米。
The gradient is the acceleration: 8 divided by 4, which is 2.0 metres per second squared.
斜率就是加速度：8除以4，等于每秒平方2.0米。
Velocity falls uniformly from +6 to −2 m/s in 4 s.
速度在 4 秒内从正 6 米每秒均匀下降至负 2 米每秒。
Find acceleration, displacement and distance.
求加速度、位移和路程。
Explain why the last two differ.
解释后两者为何不同。
The acceleration is minus 2 metres per second squared, and the displacement is 6 minus 2, times 4, over 2, which is 8 m.
加速度是每二次方秒 −2 米；位移等于 6 减 2，乘以 4，再除以 2，即 8 米。
The velocity crosses zero at 3 s, so the two triangle areas give a distance of 9 plus 1, which is 10 m.
速度在 3 秒时变为零，两个三角形的面积给出路程 9 加 1，即 10 米。
Distance adds path lengths despite the reversal.
尽管运动反向，路程仍是各段路径长度之和。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。