The diagram shows the speed-time graph for Sriya's 100 metre race during her school's sports day. In the first 5 seconds, Sriya's accelerated uniformly to a speed of 9.6 m/s. She maintained her speed for the next 6 seconds and slowed down over the last 2.2 seconds. She crossed the finishing line after 13.2 seconds.

1. Calculate the constant acceleration during the first 5 seconds

Since the acceleration is uniform from $t=0$ to $t=5$,

So, 3 seconds after the race started, the acceleration is $1.92\text{ m/s}^2$.

1. Find the distance covered in the first 11 seconds from the area under the speed-time graph

Distance in first 5 s:
$\frac{1}{2}\times 5\times 9.6=24\text{ m}$
Distance from 5 s to 11 s:
$6\times 9.6=57.6\text{ m}$
Total distance in first 11 s:
$24+57.6=81.6\text{ m}$

2. Use the remaining distance as the trapezium area to find the finishing speed

Remaining distance:
$100-81.6=18.4\text{ m}$
Let the speed at the finishing line be $v\text{ m/s}$.

For the last 2.2 s, the area under the graph is a trapezium:
$\frac{1}{2}(9.6+v)(2.2)=18.4$
$(9.6+v)(1.1)=18.4$
$9.6+v=\frac{18.4}{1.1}=16.727\ldots$
$v=16.727\ldots-9.6=7.127\ldots$
$v=7.13\text{ m/s}$

1. Find the remaining distance Ella had to run after the first 6 seconds

Distance remaining after $6$ s:
$100-29.7=70.3\text{ m}$

2. Find the time Ella took to cover the remaining distance at constant speed

After $6$ s, Ella ran at a constant speed of $10.2\text{ m/s}$.

$\text{time}=\frac{\text{distance}}{\text{speed}}=\frac{70.3}{10.2}=6.892\ldots\text{ s}$

3. Find Ella's total finishing time and compare with Sriya's time

Ella's total time:
$6+6.892\ldots=12.892\ldots\text{ s}$

Sriya's time was $13.2\text{ s}$.

Since $12.892\ldots < 13.2$, Ella finished first.