The diagram shows the speed-time graph of a motorcyclist.
The shaded area represents the distance travelled.
The distance travelled is 450 m.

1. Express the total distance as the area under the speed-time graph in terms of $v$.

Distance travelled = area under the speed-time graph

$450=\frac12(10)(v)+(30-10)(v)+\frac12(55-30)(v)$

$450=5v+20v+12.5v$

2. Solve the resulting linear equation for $v$.

$450=37.5v$

$v=\frac{450}{37.5}=12$

1. Find the acceleration during the first 10 seconds from the gradient of the speed-time graph.

From part (a), $v=12$.\n\nFor $0\leq t\leq 10$, the speed increases uniformly from $0$ to $12\text{ m/s}$.\n\n$\text{acceleration}=\frac{12-0}{10-0}=1.2\text{ m/s}^2$

2. Use the acceleration to find the speed after 8 seconds.

After $8$ seconds,\n\n$\text{speed}=1.2\times 8=9.6\text{ m/s}$

1. Interpret the horizontal part of the speed-time graph from 10 s to 30 s.

From $t=10$ to $t=30$, the speed-time graph is horizontal at speed $v$. Since $v=12$, the speed is constant at $12\text{ m/s}$.

1. Find the decrease in speed per second over the last 25 seconds.

For the last $25$ seconds, the speed decreases from $12\text{ m/s}$ to $0\text{ m/s}$.

$\text{deceleration}=\frac{12-0}{25}=0.48\text{ m/s}^2$

1. Find the distance travelled by 10 seconds.

From earlier parts, $v=12\text{ m/s}$.

Distance travelled from $0$ to $10$ s is the area of the triangle under the speed-time graph:

$\text{distance}=\frac12(10)(12)=60\text{ m}$

So the distance-time graph passes through $(10,60)$.

2. Find the distance travelled by 30 seconds.

From $10$ to $30$ s, the speed is constant at $12\text{ m/s}$.

Distance travelled in this interval:

$20\times 12=240\text{ m}$

Total distance by $30$ s:

$60+240=300\text{ m}$

So the graph passes through $(30,300)$.

3. Find the final point and choose the correct shapes of the graph.

The total distance travelled is $450\text{ m}$, and the journey ends at $55$ s.

So the final point is $(55,450)$.

For the shape:

• From $0$ to $10$ s, speed increases uniformly, so the distance-time graph has increasing gradient and is concave upwards.
• From $10$ to $30$ s, speed is constant, so the distance-time graph is a straight line.
• From $30$ to $55$ s, speed decreases uniformly to $0$, so the distance-time graph has decreasing gradient and is concave downwards, ending horizontal at $55$ s.