Describing motion · 描述运动
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| distance/ˈdɪstəns/ | 距离 | jù lí |
| displacement/dɪˈspleɪsmənt/ | 位移 | wèiyí |
| speed/spiːd/ | 速率 | sùlǜ |
| velocity/vəˈlɒsɪti/ | 速度 | sùdù |
| acceleration/əkˌseləˈreɪʃn/ | 加速度 | jiāsùdù |
| gradient/ˈɡreɪdɪənt/ | 斜率 | xié lǜ |
| deceleration/dɪˌseləˈreɪʃn/ | 减速度 | jiǎn sù dù |
| tangent/ˈtændʒənt/ | 切线 | qiè xiàn |
There and back
- A runner does one lap of a $400\ \text{m}$ track and stops where they started.
- Distance 距离 travelled: $400\ \text{m}$. Displacement 位移: $0$.
- Describing motion well starts with getting words like these exactly right.
来回一趟
- 一个跑步者在 $400\ \text{m}$ 跑道上跑一圈,停在出发的地方。
- 距离(distance):$400\ \text{m}$。位移(displacement):$0$。
- 描述运动,先从把这些词用得精确开始。
Five key words
- distance (scalar) — total path length; displacement (vector) — straight line, start to end.
- speed 速率 (scalar) — rate of change of distance; velocity 速度 (vector) — rate of change of displacement.
- acceleration 加速度 (vector) — rate of change of velocity. Units: $\dfrac{\text{m}}{\text{s}}$ and $\dfrac{\text{m}}{\text{s}^2}$.
Velocity–time graph — gradient 斜率 gives acceleration, area gives displacement
五个关键词
- 距离(标量)——总路径长度;位移(矢量)——从起点到终点的直线。
- 速率(speed)(标量)——距离的变化率;速度(velocity)(矢量)——位移的变化率。
- 加速度(acceleration)(矢量)——速度的变化率。单位:$\dfrac{\text{m}}{\text{s}}$ 和 $\dfrac{\text{m}}{\text{s}^2}$。
速度–时间图——斜率给出加速度,面积给出位移
The velocity–time graph · 速度–时间图
v = u + at
On a speed–time graph the gradient is the acceleration and the area · 面积 underneath is the distance travelled. · 在速率–时间图上,斜率 是加速度,下方的 面积 是走过的距离。
The rate of change of velocity with time is called ____. · 速度随时间的变化率叫做 ____。
Acceleration is how fast the velocity changes — a vector, in $\dfrac{\text{m}}{\text{s}^2}$. · 加速度是速度变化的快慢——一个矢量,单位 $\dfrac{\text{m}}{\text{s}^2}$。
Which quantity is the rate of change of displacement? · 哪个量是 位移的变化率?
Velocity is the rate of change of displacement · 位移 (a vector). Speed is the rate of change of distance · 距离 (a scalar). · 速度是位移的变化率(矢量)。速率是距离的变化率(标量)。
Match each term to the definition the examiner marks. · 把每个术语与评分认可的定义配对。
Speed and velocity differ by one word, distance against displacement, and that word is the mark. · 速率与速度只差一个词,路程还是位移,而那个词就是分。
Deceleration 减速度 is not a new quantity — it is just acceleration pointing opposite to the velocity (a negative acceleration).
Experimental set-up for measuring the acceleration due to free fall
减速(deceleration) 不是一个新的量——它只是方向与速度相反的加速度(一个负的加速度)。
测量自由落体加速度的实验装置
Deceleration is a completely separate quantity from acceleration. · 减速是与加速度完全不同的另一个量。
No — deceleration is just acceleration in the opposite direction to the motion (a negative acceleration). · 不是——减速只是方向与运动相反的加速度(一个负的加速度)。
Two graphs do the work
- Most motion questions use a displacement–time or a velocity–time graph.
- The whole skill is knowing what the gradient and the area mean on each.
两种图就够用了
- 大多数运动问题用 位移–时间 图或 速度–时间 图。
- 整个技巧就是知道每种图上 斜率 和 面积 表示什么。
Displacement–time graph
- The gradient (steepness) = the velocity.
- Flat → at rest. Straight → constant velocity. Curved → take a tangent 切线.
位移–时间图
- 斜率(陡峭程度)= 速度。
- 水平 → 静止。直线 → 匀速。曲线 → 作 切线。
On a displacement–time graph, a flat (horizontal) line means the object is at rest. · 在位移–时间图上,一条水平线表示物体静止。
A flat line has zero gradient, and the gradient is the velocity — so the velocity is zero (at rest). · 水平线斜率为零,而斜率就是速度——所以速度为零(静止)。
Velocity–time graph
- The gradient = the acceleration.
- The area under the line = the displacement.
速度–时间图
- 斜率 = 加速度。
- 线下的 面积 = 位移。
Match each graph feature to what it tells you. · 把每个图象特征与它告诉你的信息配对。
Gradients give rates of change; the area under a velocity–time graph builds up the displacement. · 斜率给出变化率;速度–时间图线下的面积累积成位移。
Displacement from the area
- Split the area into triangles and rectangles, then add them up.
- Triangle $= \tfrac{1}{2} \times \text{base} \times \text{height}$; rectangle $= \text{base} \times \text{height}$.
由面积求位移
- 把面积分成 三角形和矩形,再加起来。
- 三角形 = ½ × 底 × 高;矩形 = 底 × 高。
A car speeds up from rest to $20\ \dfrac{\text{m}}{\text{s}}$ in · 入 $10\ \text{s}$ (a straight velocity–time line). How far does it travel? · 一辆车从静止在 $10\ \text{s}$ 内加速到 $20\ \dfrac{\text{m}}{\text{s}}$(速度–时间图是一条直线)。它走了多远?
Displacement = area under the line = $\tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 10 \times 20 = 100\ \text{m}$. · 位移 = 线下面积 = ½ × 底 × 高 = $\tfrac{1}{2} \times 10 \times 20 = 100\ \text{m}$。
Worked example: reading a lift's journey
- A velocity-time graph shows a lift rising: $0$ to $2.0\ \text{m/s}$ in $4.0\ \text{s}$, then steady for $10\ \text{s}$, then back to rest in $2.0\ \text{s}$. Find the acceleration in each stage and the total distance.
- Gradients give the accelerations: $2.0/4.0 = 0.50\ \text{m/s}^2$, then $0$, then $-2.0/2.0 = -1.0\ \text{m/s}^2$.
- Areas give the distances: a triangle $\tfrac{1}{2}(4.0)(2.0) = 4.0\ \text{m}$, a rectangle $(10)(2.0) = 20\ \text{m}$, a triangle $\tfrac{1}{2}(2.0)(2.0) = 2.0\ \text{m}$.
- Total: $26\ \text{m}$. Split the graph into triangles and rectangles and add. Never try to fit one formula to the whole journey.
- The commonest error here is taking the gradient when the question wants the area. Gradient is acceleration; area is displacement.
例题:读懂一部电梯的行程
- 一张速度-时间图显示电梯上行:$4.0\ \text{s}$ 内从 $0$ 到 $2.0\ \text{m/s}$,然后匀速 $10\ \text{s}$,再用 $2.0\ \text{s}$ 回到静止。求各阶段的加速度和总路程。
- 斜率给出加速度:$2.0/4.0 = 0.50\ \text{m/s}^2$,然后是 $0$,再是 $-2.0/2.0 = -1.0\ \text{m/s}^2$。
- 面积给出路程:一个三角形 $\tfrac{1}{2}(4.0)(2.0) = 4.0\ \text{m}$、一个矩形 $(10)(2.0) = 20\ \text{m}$、一个三角形 $\tfrac{1}{2}(2.0)(2.0) = 2.0\ \text{m}$。
- 总计 $26\ \text{m}$。把图分成三角形和矩形再相加。绝不要用一个公式去套整段行程。
- 这里最常见的错误是题目要面积时却去求斜率。斜率是加速度;面积是位移。
On a velocity-time graph a lift speeds up from rest to 2.0 m/s in 4.0 s. How far does it travel in that stage, in metres? · 在速度-时间图上,一部电梯在 4.0 s 内从静止加速到 2.0 m/s。这一阶段它走了多少米?
The area under that part is a triangle: half of 4.0 x 2.0 = 4.0 m. Area is displacement; the gradient, 0.50 m/s^2, is the acceleration. · 那一段下面的面积是一个三角形:4.0 x 2.0 的一半 = 4.0 m。面积是位移;斜率 0.50 m/s^2 才是加速度。
Worked example: when there is no time given
- A car accelerates uniformly from $8.0\ \text{m/s}$ to $20\ \text{m/s}$ over $56\ \text{m}$. Find its acceleration.
- List what you have: $u = 8.0$, $v = 20$, $s = 56$, $a = ?$, and $t$ is not wanted and not given.
- So choose the equation with no $t$ in it: $v^2 = u^2 + 2as$.
- $20^2 = 8.0^2 + 2a(56)$, so $400 - 64 = 112a$ and $a = 3.0\ \text{m/s}^2$.
- That is the whole method for these questions: write out $s, u, v, a, t$, mark the one you do not have and do not want, and pick the equation that is missing it.
例题:题目没给时间的时候
- 一辆车在 $56\ \text{m}$ 内从 $8.0\ \text{m/s}$ 匀加速到 $20\ \text{m/s}$。求它的加速度。
- 列出已知:$u = 8.0$、$v = 20$、$s = 56$、$a = ?$,而 $t$ 既不要求也没给。
- 所以选那个不含 $t$ 的方程:$v^2 = u^2 + 2as$。
- $20^2 = 8.0^2 + 2a(56)$,于是 $400 - 64 = 112a$,$a = 3.0\ \text{m/s}^2$。
- 这类题的方法就是这样:把 $s, u, v, a, t$ 写出来,标出那个既没有又不想要的量,挑那个不含它的方程。
Put the method for a constant-acceleration problem in order. · 把恒定加速度问题的做法按顺序排列。
Choosing the equation that is missing the quantity you neither have nor want avoids solving two equations at once. · 挑那个不含你既没有又不想要的量的方程,可以省掉联立两个方程。
A car accelerates uniformly from 8.0 m/s to 20 m/s over 56 m. What is its acceleration, in m/s^2? · 一辆车在 56 m 内从 8.0 m/s 匀加速到 20 m/s。它的加速度是多少 m/s^2?
No time is given or wanted, so use v^2 = u^2 + 2as: 400 - 64 = 112a, a = 3.0 m/s^2. · 时间既没给也不要,所以用 v^2 = u^2 + 2as:400 - 64 = 112a,a = 3.0 m/s^2。
Marks that slip away
- Gradient is acceleration, area is displacement. Read which one the question wants before touching the graph.
- The equations of motion apply only to constant acceleration. On a curved graph only the graph methods work.
- Choose one direction as positive and keep the signs consistent for the whole journey. With up positive, $a = -9.81\ \text{m/s}^2$ on the way down as well as on the way up.
- Distance is the total path length; displacement is from start to finish in a stated direction. A there-and-back trip has a distance but zero displacement.
- Speed is the rate of change of distance, velocity the rate of change of displacement. The two definitions differ by one word and that word is the mark.
容易丢掉的分
- **斜率是加速度,面积是位移。**动图之前先看清题目要的是哪一个。
- 运动学方程只适用于恒定加速度。图是曲线时,只有图解法管用。
- 选定一个方向为正,并在整段行程中保持符号一致。以向上为正时,下落途中的 $a$ 也是 $-9.81\ \text{m/s}^2$。
- 路程是走过的全部路径长度;位移是从起点到终点、带方向的。往返一趟有路程而位移为零。
- 速率是路程的变化率,速度是位移的变化率。两个定义只差一个词,而那个词就是分。
Which of these are errors in kinematics questions? Select all · 所有 that apply. · 下列哪些是运动学题中的错误?选出所有适用的。
The last one is the correct method. The sign error is the subtlest: once up is positive, a stays negative for the whole flight, downward part included. · 最后一条正是正确做法。符号错误最隐蔽:一旦取向上为正,整段飞行中 a 都保持为负,下落部分也不例外。
You've got it
- distance & speed are scalars; displacement, velocity & acceleration are vectors
- displacement–time gradient = velocity; velocity–time gradient = acceleration
- area under a velocity–time graph = the displacement
你掌握了
- 距离和速率是 标量;位移、速度和加速度是 矢量
- 位移–时间图 斜率 = 速度;速度–时间图 斜率 = 加速度
- 速度–时间图线下的 面积 = 位移