该图显示了一个质点的速度-时间图象。

1. Find the increase in speed per second on the straight-line section from $6$ s to $13$ s.

From $t=6$ to $t=13$, the speed increases from $30$ m/s to $100$ m/s.

$\text{Increase per second}=\frac{100-30}{13-6}=\frac{70}{7}=10\text{ m/s per s}$

2. Use the linear increase to find the speed at $t=12$ s.

At $t=12$, the time elapsed after $t=6$ is

$12-6=6\text{ s}$

So the speed is

$30+10(6)=90\text{ m/s}$

1. Identify the required region under the speed-time graph before slowing down.

The particle starts to slow down at $t=13$ s.

Distance covered before this is the area under the speed-time graph from $t=0$ to $t=13$ s.

2. Find the distance from $0$ s to $6$ s using the rectangular area.

From $0$ to $6$ s, speed is constant at $30$ m/s.

$\text{Distance}=6\times 30=180\text{ m}$

3. Find the distance from $6$ s to $13$ s using the trapezium area, then add the distances.

From $6$ to $13$ s, the speeds are $30$ m/s and $100$ m/s.

$\text{Distance}=\frac{1}{2}(30+100)(13-6)$
$=\frac{1}{2}(130)(7)=455\text{ m}$

Total distance before slowing down:

$180+455=635\text{ m}$

1. Identify the key distances and use the speeds to decide the shape of the distance-time graph.

From earlier parts,
$\text{distance at }t=6=30\times 6=180\text{ m}$
$\text{distance at }t=13=635\text{ m}$
From $13$ s to $21$ s, the extra distance is the triangular area under the speed-time graph:
$\frac12\times 8\times 100=400\text{ m}$
So
$\text{distance at }t=21=635+400=1035\text{ m}.$
On the distance-time graph, draw through these positions if the scale allows. The gradient represents speed, so:

• $0\leq t\leq 6$: constant speed, so draw a straight line.
• $6\leq t\leq 13$: speed increases, so draw a curve getting steeper.
• $13\leq t\leq 21$: speed decreases to $0$, so draw a curve getting less steep, ending horizontally.