两列火车 $A$ 和 $B$ 在 $t=0$ 时从同一车站同时由静止出发，沿平行的水平直轨道行驶。

两列火车均在 $180\,\mathrm{s}$ 后到达下一站并静止。

火车 $A$ 先以恒定加速度 $\frac23\,\mathrm{m\,s^{-2}}$ 行驶 $30\,\mathrm{s}$，再以恒定速率行驶 $120\,\mathrm{s}$，最后以恒定减速度行驶 $30\,\mathrm{s}$。

火车 $B$ 先以恒定加速度行驶 $90\,\mathrm{s}$，再在最后 $90\,\mathrm{s}$ 内以恒定减速度行驶。

(a) 在同一坐标系中画出两列火车在两站之间运动的速率—时间图像草图。

(b) 求火车 $B$ 在行程前半段的加速度。

(c) 求两列火车速率相同的时刻。

(d) 求出发 $96\,\mathrm{s}$ 后两列火车之间的距离。

$A$ 的速率图为等腰梯形，$B$ 的图为等腰三角形。两车同起点同终点且同向，图下面积相等，因此 $B$ 的最高速率大于 $A$，两图在行驶中相交两次。

Draw both graphs on the same axes, with time $t\text{ (s)}$ horizontally and speed $v\text{ (m s}^{-1}\text{)}$ vertically.

For $A$, the maximum speed is

Join $(0,0)$, $(30,20)$, $(150,20)$ and $(180,0)$ by straight lines.

For $B$, draw an isosceles triangle from $(0,0)$ to a peak at $t=90$, then to $(180,0)$. Its peak must be above $20\text{ m s}^{-1}$, so the graphs cross twice between the stations. Label the two trains and the times $30,90,150,180$.

两车经过的总路程相同，用图下面积先求站间距离。$B$ 在前 $90$ 秒达到最高速率，再由速度增量除以时间求加速度。

The distance between the stations is the area under $A$‘s graph:

Let the maximum speed of $B$ be $V$. Equating the area under its graph to this distance,

Therefore the acceleration of $B$ is

行驶中速率相等的两个交点都在 $A$ 匀速阶段。先分别写出 $B$ 加速段和减速段的速率，解后检查时间所在区间。

During $30\leq t\leq150$, train $A$ has speed $20\text{ m s}^{-1}$.

On $B$‘s accelerating section,

This lies in $30<t<90$, as required.

On $B$‘s decelerating section,

This lies in $90<t<150$. Thus, while the trains are moving, the required times are

在 $t=90$ 时，两车均恰好走过一半路程，因此位置相同。只比较接下来 $6$ 秒的位移，可减少计算。

By symmetry of each speed–time graph, both trains have travelled $1500\text{ m}$ at $t=90$.

From $t=90$ to $t=96$, train $A$ travels

Train $B$ has initial speed $100/3\text{ m s}^{-1}$ and acceleration $-10/27\text{ m s}^{-2}$. Its displacement in these $6$ seconds is

Therefore $B$ is ahead of $A$ by

也可从终点往回计算剩余路程。两车终点相同，剩余路程之差就是此时的间距。

At $t=96$, train $B$ has $180-96=84$ seconds remaining. Its remaining distance is the area of its final triangular section:

Train $A$ has $54$ seconds at speed $20$ followed by its final $30$-second deceleration. Its remaining distance is

Thus $B$ is ahead, and the separation is