📚 Mathematics of Motion: Speed, Time and Distance | 数学行程问题：速度、时间与路程的关系
Every moving object — a car on a highway, a runner on a track, or water flowing in a river — follows a fundamental relationship: distance equals speed multiplied by time. Mastering this relationship unlocks the ability to solve real-world motion problems efficiently and accurately.
每一个运动的物体——无论是高速公路上的汽车、跑道上的运动员，还是河流中流淌的水——都遵循一个基本关系：路程等于速度乘以时间。掌握这一关系，能够让我们高效且准确地解决现实中的行程问题。
1. The Golden Formula | 黄金公式
The core equation is simple but powerful: \( \text{Distance} = \text{Speed} \times \text{Time} \). From this, we derive two equivalent forms: Speed = Distance ÷ Time, and Time = Distance ÷ Speed. These three are interchangeable, and choosing the right one depends on which two variables you know.
核心公式简洁而有力：路程 = 速度 × 时间。由此可推导出两个等价形式：速度 = 路程 ÷ 时间，以及时间 = 路程 ÷ 速度。这三个公式可以互相转换，选择哪一个取决于你已知的是哪两个变量。
Distance = Speed × Time
Speed = Distance ÷ Time
Time = Distance ÷ Speed
Always ensure the units are consistent before substituting values. If speed is in kilometres per hour and time is in hours, the distance will be in kilometres. Mixing units is the most common source of error in these problems.
在代入数值之前，务必确保单位一致。如果速度以千米每小时为单位，时间以小时为单位，那么路程的单位就是千米。单位混用是此类问题中最常见的错误来源。
2. Unit Conversions | 单位换算
Examiners frequently test unit conversion skills within motion problems. Converting between m/s and km/h is especially common. The factor is 3.6 — to convert from m/s to km/h, multiply by 3.6; to convert from km/h to m/s, divide by 3.6.
考官经常在行程问题中考查单位换算能力。米每秒与千米每小时之间的转换尤为常见。换算系数是3.6——将米每秒转换为千米每小时，乘以3.6；将千米每小时转换为米每秒，除以3.6。
1 m/s = 3.6 km/h
This arises because 1 km = 1000 m and 1 hour = 3600 s, so 1 km/h = 1000 m ÷ 3600 s = ¹⁰⁄₃₆ m/s = ⁵⁄₁₈ m/s, and reciprocally 1 m/s = ¹⁸⁄₅ km/h = 3.6 km/h.
这个换算关系来源于：1千米 = 1000米，1小时 = 3600秒，因此 1千米/小时 = 1000米 ÷ 3600秒 = ¹⁰⁄₃₆ 米/秒 = ⁵⁄₁₈ 米/秒，反过来 1米/秒 = ¹⁸⁄₅ 千米/小时 = 3.6 千米/小时。
| Speed (m/s) | Speed (km/h) |
|---|---|
| 5 | 18 |
| 10 | 36 |
| 15 | 54 |
| 20 | 72 |
When converting speeds involving minutes instead of hours, remember to convert time into the same base unit. For example, 20 m/s for 2 minutes = 20 × 120 = 2400 m = 2.4 km.
当涉及以分钟为单位的时间而速度以米每秒为单位时，记得将时间转换为相同的基本单位。例如，以20米/秒的速度行驶2分钟，则距离 = 20 × 120 = 2400米 = 2.4千米。
3. Average Speed | 平均速度
Average speed is not simply the arithmetic mean of two speeds. It is defined as total distance travelled divided by total time elapsed. This distinction is critical when a journey consists of multiple legs at different speeds.
平均速度并不是两个速度的算术平均值。它定义为总路程除以总时间。当旅程包含多个不同速度的路段时，这一区分至关重要。
Average Speed = Total Distance ÷ Total Time
For example, a car travels 100 km at 50 km/h and then another 100 km at 100 km/h. The arithmetic mean is 75 km/h, but the actual average speed is 200 km ÷ (2 h + 1 h) = 200 ÷ 3 ≈ 66.7 km/h. The slower leg occupies more time, pulling the average downward.
例如，一辆汽车以50公里/小时行驶100公里，再以100公里/小时行驶100公里。算术平均值是75公里/小时，但实际平均速度为200公里 ÷ (2小时 + 1小时) = 200 ÷ 3 ≈ 66.7公里/小时。较慢的路段占用了更多时间，从而拉低了平均值。
A common exam trap is the “there and back” problem with equal distances. If a cyclist travels from A to B at speed v₁ and returns at speed v₂, the average speed for the round trip is 2v₁v₂ ÷ (v₁ + v₂), not (v₁ + v₂) ÷ 2.
一个常见的考题陷阱是等距离往返问题。如果骑行者从A到B的速度为v₁，返回速度为v₂，那么整个往返的平均速度为 2v₁v₂ ÷ (v₁ + v₂)，而不是 (v₁ + v₂) ÷ 2。
4. Meeting and Overtaking | 相遇与追及
When two objects move towards each other, their relative speed is the sum of their speeds: v₁ + v₂. When they move in the same direction, the relative speed is the difference: |v₁ − v₂|. Relative speed is the rate at which the distance between them changes.
当两个物体相向运动时，它们的相对速度是各自速度之和：v₁ + v₂。当它们同向运动时，相对速度是速度之差：|v₁ − v₂|。相对速度是它们之间距离变化的速率。
Time to Meet = Initial Distance ÷ (v₁ + v₂)
Time to Overtake = Initial Gap ÷ (v₁ − v₂)
Consider two trains facing each other from stations 300 km apart. Train A runs at 80 km/h, Train B at 70 km/h. Their closing speed is 150 km/h, so they meet after 300 ÷ 150 = 2 hours. If the same two trains travel in the same direction with Train A behind, the gap closes at only 10 km/h, taking 30 hours.
考虑两列相向行驶的火车，从相距300公里的两个车站出发。甲车速度为80公里/小时，乙车速度为70公里/小时。它们的接近速度为150公里/小时，因此2小时后相遇（300 ÷ 150 = 2小时）。如果两列火车同向行驶且甲车在后，间距以10公里/小时的速度缩小，则需要30小时才能追上。
5. Circular Track Problems | 环形跑道问题
Circular track problems are a classic extension of meeting and overtaking. When two runners start from the same point on a circular track of circumference C and move in the same direction, the faster runner gains one full lap each time the distance between them equals C. Thus:
环形跑道问题是相遇与追及问题的经典延伸。当两名运动员从环形跑道（周长C）上同一点同向出发时，每次两者之间的距离等于C时，较快的运动员就领先一圈。因此：
First Overtake Time = C ÷ (v₁ − v₂)
If they move in opposite directions, each time they meet corresponds to the combined distance covering exactly one full circumference:
如果他们相向运动，每次相遇对应着两人合起来跑完一整圈周长：
First Meeting Time = C ÷ (v₁ + v₂)
For a track of 400 m with runners at 6 m/s and 4 m/s running in the same direction, the faster runner laps the slower one every 400 ÷ (6 − 4) = 200 seconds. Running in opposite directions, they meet every 400 ÷ (6 + 4) = 40 seconds.
对于周长400米的跑道，两位跑者分别以6米/秒和4米/秒同向奔跑，速度较快者每隔 400 ÷ (6 − 4) = 200秒 套圈一次。若相向奔跑，则每隔 400 ÷ (6 + 4) = 40秒 相遇一次。
6. Effect of Current: Boat and River | 流水行船问题
For a boat moving through a river, the current either assists or opposes its motion. Let the boat’s speed in still water be v and the current speed be u. Downstream speed is v + u; upstream speed is v − u. This mirrors the relative speed principle applied to a moving medium.
对于在河流中行驶的船只，水流要么助力、要么阻碍其运动。设船在静水中的速度为v，水流速度为u。顺水速度为v + u；逆水速度为v − u。这体现了相对速度原理在运动介质中的应用。
Downstream speed = v + u; Upstream speed = v − u
A boat takes 4 hours to travel 60 km downstream and 6 hours for the same distance upstream. Then v + u = 15 km/h and v − u = 10 km/h. Solving simultaneously: v = 12.5 km/h, u = 2.5 km/h.
一艘船顺水60公里需4小时，逆水同样60公里需6小时。则 v + u = 15公里/小时，v − u = 10公里/小时。联立解方程组：v = 12.5公里/小时，u = 2.5公里/小时。
When the boat turns around mid-journey or the current reverses direction, carefully track the sign of each leg. Always draw a diagram or write a timeline to avoid confusion.
当船中途掉头或水流方向改变时，务必仔细追踪每一段航程中速度的正负号。建议画图或列出时间线以避免混淆。
7. Simultaneous Motion from Two Points | 两地同时出发问题
When two objects depart from different points at different times, the key is to align their starting times. Suppose Alice leaves A at 2:00 pm at 60 km/h, and Bob leaves B at 3:00 pm towards A at 40 km/h. The distance between A and B is 200 km.
当两个物体从不同地点在不同时间出发时，关键在于对齐它们的起始时间。假设爱丽丝下午2:00从A地以60公里/小时出发，鲍勃下午3:00从B地向A地以40公里/小时出发。A、B两地相距200公里。
By 3:00 pm, Alice has already covered 60 km, so the remaining distance between them is 140 km. Their combined speed is 100 km/h, so they meet at 3:00 + 1.4 h = 4:24 pm. Alice’s total distance is 60 + 60 × 1.4 = 144 km from A.
到下午3:00时，爱丽丝已行驶60公里，因此两人之间的距离为140公里。他们的合速度为100公里/小时，所以他们相遇于 3:00 + 1.4小时 = 下午4:24。爱丽丝从A地出发的总行驶距离为 60 + 60 × 1.4 = 144公里。
Breaking the problem into time segments — before Q departs and after Q departs — simplifies computation. Never add distances at different reference frames without alignment.
将问题按时间分段——先行驶段与后出发段——可以简化计算。切勿在未对齐参考时间的情况下直接相加距离。
8. Graphical Interpretation | 图形解读
A distance–time graph plots distance on the vertical axis and time on the horizontal axis. The slope at any point equals the instantaneous speed. A horizontal segment indicates a stop; a steep segment indicates rapid motion. In a speed–time graph, the area under the curve equals the total distance travelled.
路程-时间图以纵轴为路程、横轴为时间。任意点的斜率等于瞬时速度。水平段表示停止，陡峭段表示快速运动。在速度-时间图中，曲线下方的面积等于总路程。
Area under speed–time graph = Total distance
For a body accelerating uniformly from u to v over time t, the distance is the area of a trapezium: s = (u + v) × t ÷ 2. This coincides with s = ut + ½at², the kinematic equation for uniform acceleration.
对于从初速度u匀加速到末速度v、历时t的物体，路程是梯形的面积：s = (u + v) × t ÷ 2。这与匀加速运动学方程 s = ut + ½at² 一致。
When solving graph-based questions, annotate the graph with known values, break irregular shapes into rectangles and triangles, and double-check the units of the axes before computing areas.
在解答图形类题目时，在图上标注已知数值，将不规则图形拆分为矩形和三角形，并在计算面积前仔细检查坐标轴的单位。
9. Problems with Stops: Effective Speed | 含停顿问题：有效速度
A car travels at 60 km/h but stops for 10 minutes every hour. The naive approach is to use 60 km/h for the entire journey, but the effective speed is lower. Over one full hour including a stop, the travel time is only 50 minutes, so the distance covered is 60 km/h × ⁵⁰⁄₆₀ h = 50 km. The effective speed is 50 km/h.
一辆汽车以60公里/小时行驶，但每小时停车10分钟。直观上直接用60公里/小时计算全程是错误的，有效速度其实更低。在一个包含停车的完整小时内，实际行驶时间只有50分钟，所以行驶距离为 60公里/小时 × ⁵⁰⁄₆₀ 小时 = 50公里。有效速度为50公里/小时。
Effective Speed = Distance ÷ (Moving Time + Stopped Time)
If a journey of 150 km includes four 5-minute fuel stops, first calculate the pure driving time (150 ÷ 60 = 2.5 h), add the total stop time (4 × 5 min = 20 min = ⅓ h), yielding a total journey time of 2.833 h. The effective speed is then 150 ÷ 2.833 ≈ 52.9 km/h.
如果一次150公里的旅程包含四次各5分钟的加油停靠，先计算纯驾驶时间（150 ÷ 60 = 2.5小时），再加上总停车时间（4 × 5分钟 = 20分钟 = ⅓小时），得到总行程时间为2.833小时。那么有效速度为 150 ÷ 2.833 ≈ 52.9公里/小时。
10. Ratio Methods in Distance Problems | 距离问题中的比例法
Many motion problems become simpler with ratios. When the distance is fixed, speed and time are inversely proportional. If speed increases by a factor of 3, time decreases to one-third. When the time is fixed, distance is directly proportional to speed.
许多行程问题用比例法会更简单。当距离固定时，速度与时间成反比。如果速度增加到原来的3倍，时间就减少到原来的三分之一。当时间固定时，路程与速度成正比。
If distance fixed: v₁ × t₁ = v₂ × t₂
A train normally covers a route at 80 km/h taking 6 hours. After track upgrades it travels at 100 km/h. The new time is 80 × 6 ÷ 100 = 4.8 hours, saving 1.2 hours. This proportional shortcut avoids computing the actual distance.
一列火车通常以80公里/小时的速度行驶一条线路，需要6小时。轨道升级后以100公里/小时行驶。新时间为 80 × 6 ÷ 100 = 4.8小时，节省了1.2小时。这种比例快捷方法无需计算实际距离。
For two moving objects starting simultaneously from opposite ends, the ratio of distances travelled at the meeting point equals the ratio of their speeds. This is a powerful tool for geometry-style distance questions.
对于从两端同时相向出发的两个运动物体，相遇时各自行驶的路程之比等于它们的速度之比。这是解决几何风格距离问题的利器。
11. Common Pitfalls | 常见错误防范
Students commonly make four errors: (1) using average of speeds instead of total distance ÷ total time; (2) forgetting to convert units before computation; (3) adding distances wrong when starting times differ; (4) confusing relative speed for opposite and same direction cases.
学生常犯四类错误：（1）直接用速度平均值替代总路程÷总时间；（2）计算前忘记统一单位；（3）出发时间不同时错误地直接相加路程；（4）混淆相向与同向时相对速度的计算方式。
To avoid these, follow a disciplined routine: write down known quantities, underline the unknown, draw a simple diagram, label all units, and verify the result with a sanity check — is the answer physically plausible? A car cannot take 0.3 hours to travel 200 km unless its speed is absurdly high.
为避免这些错误，养成规范解题习惯：列出已知量、标出未知量、画简单示意图、标注所有单位，并用合理性质疑验证结果——物理上是否合理？一辆汽车不可能只用0.3小时行驶200公里，除非速度高得离谱。
Additionally, when reading graphs, always note the exact scale: one small grid square may equal 5 km or 0.5 hours. Misreading the scale is a silent killer in exam settings.
此外，读图时必须注意刻度：一个小方格可能代表5公里或0.5小时。误读刻度是考试中不易察觉的致命错误。
12. Exam Strategy and Practice Tips | 考试策略与练习建议
In a timed exam, allocate roughly one minute per mark. For a 4-mark question, spend no more than four minutes. If a problem seems intractable, write the formula for distance, speed, or time, substitute known values, and see if partial credit can be secured.
在计时考试中，大约为每分值分配一分钟。对于4分值的题目，花费不超过4分钟。如果题目看起来棘手，先写出路程、速度或时间的公式，代入已知数值，看看能否获得部分步骤分。
Practice with a structured approach: solve at least five problems per topic type — basic formula, average speed, relative motion, circular track, and boat-and-river. Keep an error log. Revisit the same problem a week later to reinforce memory and speed.
以系统化的方式进行练习：每种题型至少做五道——基础公式、平均速度、相对运动、环形跑道、流水行船。保持错误记录本。一周后重做同一道题，以强化记忆和速度。
Finally, memorise the key constants: 3.6 for m/s to km/h, ⁵⁄₁₈ for the reverse; the factor 60 for hours to minutes; and 60 minutes = 1 hour, 60 seconds = 1 minute. These simple anchors prevent unit chaos in high-pressure conditions.
最后，牢记关键常数：米/秒与千米/小时之间的换算系数3.6（反向为⁵⁄₁₈）；小时与分钟之间的换算系数60；以及60分钟 = 1小时，60秒 = 1分钟。这些简单的基准值能防止高压环境下的单位混乱。
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