Nurul took 2.5 hours to travel at a constant speed from Town $A$ to $B$.
On her journey back, she increased her constant speed by 4 km/h and took 15 minutes less.

1. Formulate expressions for the distance travelled in both directions

Let the speed from Town $A$ to $B$ be $v$ km/h.
Distance from $A$ to $B$ = $2.5v$
Speed from $B$ to $A$ = $(v + 4)$ km/h
Time taken from $B$ to $A$ = $2.5 - \frac{15}{60} = 2.25$ hours
Distance from $B$ to $A$ = $2.25(v + 4)$

2. Equate the distances and solve for the speed

Since the distance is the same:
$2.5v = 2.25(v + 4)$
$2.5v = 2.25v + 9$
$0.25v = 9$
$v = \frac{9}{0.25} = 36$
Therefore, her travelling speed from Town $A$ to $B$ is 36 km/h.

1. Calculate the distance between Town A and Town B

$\text{Distance} = \text{Speed} \times \text{Time} = 36 \times 2.5 = 90 \text{ km}$

2. Calculate the total time taken for the whole journey

$\text{Time for return journey} = 2.5 - \frac{15}{60} = 2.25 \text{ hours} \text{Total time} = 2.5 + 2.25 = 4.75 \text{ hours}$

3. Calculate the average speed for the whole journey

$\text{Total distance} = 90 \times 2 = 180 \text{ km} \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{180}{4.75} = 37.8947... \approx 37.9 \text{ km/h}$

1. Determine the distance from Town A to Town B

Distance $= \text{Speed} \times \text{Time} = 36 \times 2.5 = 90 \text{ km}$

2. Determine the time taken for the return journey and the total time

Time for return journey $= 2.5 \text{ h} - 15 \text{ mins} = 2.5 - 0.25 = 2.25 \text{ h}$
Total time $= 2.5 + 2.25 = 4.75 \text{ h}$

3. Identify the key points to plot on the distance-time graph

Start point: $(0, 0)$
Arrival at Town B: $(2.5, 90)$
Return to Town A: $(4.75, 0)$