The diagram shows the speed-time graph of a cyclist from Town A to Town B.

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

The distance travelled is the area under the speed-time graph.

The graph consists of two triangles and one rectangle:

$14=\frac{1}{2}\left(\frac{11}{60}\right)v+\left(\frac{42}{60}\right)v+\frac{1}{2}\left(\frac{10}{60}\right)v$

2. Solve the resulting linear equation for $v$

$14=\left(\frac{11}{120}+\frac{42}{60}+\frac{10}{120}\right)v$

$14=\left(\frac{11+84+10}{120}\right)v$

$14=\frac{105}{120}v$

$v=14\times \frac{120}{105}=16$

1. Find the speed at $t=5$ minutes using linear proportion

From part (a), $v=16$.

In the first $11$ minutes, the speed increases linearly from $0$ to $16$ km/h.

At $5$ minutes,
$\text{speed}=\frac{5}{11}\times 16=\frac{80}{11}=7.27\text{ km/h}$

1. Calculate the acceleration as change in speed divided by change in time

In the last 10 minutes, the speed decreases from $16$ km/h to $0$ km/h.

$10$ minutes $=\dfrac{10}{60}=\dfrac{1}{6}$ h.

$\text{acceleration}=\frac{0-16}{1/6}=-16\times 6=-96\text{ km/h}^2$