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. Define the unknown speed and express both journey times in hours.

Let her speed from Town A to B be $x\text{ km/h}$.

Time from A to B $=2.5$ h.

Return time $=2.5\text{ h}-15\text{ min}=2.5-0.25=2.25$ h.

Return speed $=(x+4)\text{ km/h}$.

2. Form and solve an equation using equal distances for the outward and return journeys.

Since the distance from A to B is the same as the distance from B to A,

$2.5x=2.25(x+4)$

$2.5x=2.25x+9$

$0.25x=9$

$x=36$

1. Find the total distance travelled for the return journey.

Distance from Town A to Town B:

$36 \times 2.5 = 90\text{ km}$

Total distance for the whole journey:

$90 \times 2 = 180\text{ km}$

2. Find the total time and calculate the average speed.

Return time:

$2.5\text{ h} - 15\text{ min} = 2.5\text{ h} - 0.25\text{ h} = 2.25\text{ h}$

Total time:

$2.5 + 2.25 = 4.75\text{ h}$

Average speed:

$\dfrac{180}{4.75} = 37.894\ldots$

$\approx 37.9\text{ km/h}$

1. Find the coordinates of the key turning points for the journey.

From earlier parts, the speed from Town A to B is $36\text{ km/h}$ and the time taken is $2.5\text{ h}$.

Distance from Town A to B:
$36\times 2.5=90\text{ km}$

Return speed:
$36+4=40\text{ km/h}$

Return time:
$2.5\text{ h}-15\text{ min}=2.5-0.25=2.25\text{ h}$

Total time when Nurul returns to Town A:
$2.5+2.25=4.75\text{ h}$

So the key points are
$(0,0),\quad (2.5,90),\quad (4.75,0).$

2. Plot and join the points correctly to show constant-speed travel out and back.

On the distance-time graph:

• Plot $(0,0)$.
• Plot $(2.5,90)$ for arrival at Town B.
• Plot $(4.75,0)$ for return to Town A.
• Join $(0,0)$ to $(2.5,90)$ with a straight line.
• Join $(2.5,90)$ to $(4.75,0)$ with a straight line.