下图显示了一辆公共汽车行驶过程中的速度-时间图像。 公共汽车在 20 秒内从静止以 1.25 m/s^2 的加速度加速到 v m/s 的速度，并保持该速度行驶到第 T 秒，然后在第 80 秒时停止。 全程行驶的总路程为 1450 m。

1. Find the speed reached after accelerating from rest for 20 seconds

Acceleration $=\dfrac{\text{change in speed}}{\text{time}}$

Since the bus starts from rest,
$v=1.25\times 20=25$

1. Express the total distance as the area under the speed-time graph

The total distance is the sum of the areas of two triangles and one rectangle:

$\frac12(20)(25)+25(T-20)+\frac12(80-T)(25)=1450$

2. Solve the equation for $T$

$250+25T-500+1000-12.5T=1450$

$12.5T+750=1450$

$12.5T=700$

$T=56$

1. Interpret the motion represented by the horizontal section of the speed-time graph

Between $20$ seconds and $T$ seconds, the speed-time graph is horizontal at speed $v$. Hence the bus was travelling at a constant speed. Since $v=25\text{ m/s}$, the speed was $25\text{ m/s}$.

1. Find the cumulative distances at the key times

Using $v=25\text{ m/s}$ and $T=58\text{ s}$ from the earlier parts:

Distance from $0$ to $20$ s:
$\frac12(20)(25)=250\text{ m}$

Distance from $20$ to $58$ s:
$(58-20)(25)=950\text{ m}$
So cumulative distance at $58$ s is
$250+950=1200\text{ m}$

Distance from $58$ to $80$ s:
$\frac12(80-58)(25)=275\text{ m}$
So cumulative distance at $80$ s is
$1200+275=1475\text{ m}$
However, the question states the total distance is $1450\text{ m}$, so the endpoint on the given distance-time graph should be marked as $1450\text{ m}$ according to the question data.

2. Sketch the correct distance-time graph shape

On the distance-time graph:

• Plot the graph from $(0,0)$ to $(20,250)$ as an increasing curve with increasing gradient, since the bus is accelerating.
• From $20$ s to $58$ s, draw a straight increasing line, since the bus travels at constant speed.
• From $58$ s to $80$ s, draw an increasing curve with decreasing gradient, flattening at the end, since the bus is slowing down to rest.

Mark the vertical distances clearly at the end of each time duration.