Run, mathematician, RUN!

First, let's find the distance I ran on the first 10 minutes. To be practical, I will find the average speed by basic weighted average:

$\frac{6 \times 3 + 7.5 \times 2 + 8 \times 3 + 9 \times 2}{10} = 7.5km/h$

So, the distance I ran in the first 10 minutes was:

$d = 7.5 \times \frac{10}{60}$

$d = 1.25km$

What gives me that I have to run more 1.25 km to get to 2.5km.

Let the time I ran at 5 km/h be $T_{1}$ and the time I ran at 10 km/h be $T_{2}$ .

Knowing that $T_{1} + T_{2} = 10min$ , or, for that case, $600s$ , we can write the following, bearing in mind the old formula $d = s \times t$ :

$\begin{cases} \frac { 5T_{1} }{ 3.6 } \quad +\quad \frac { 10T_{2} }{ 3.6 } \quad =\quad 1250 \\ T_{1}\quad +\quad T_{2}\quad =\quad 600 \end{cases}$ (had to divide by 3.6 to give us the answer in seconds already)

$\begin{cases} 5T_{1} \quad +\quad 10T_{2} \quad =\quad 4500 \\ T_{1}\quad +\quad T_{2}\quad =\quad 600 \quad (-10) \end{cases}$

$-5T_{1} \quad =\quad -1500$

$T_{1}\quad = \quad 300s$

Adding to the first 10 minutes, we have $\boxed{900s}$

3min at 6 Km/h = 18/60 km

2min at 7.5 Km/h = 15/60 km

3min at 8 Km/h = 24/60 km

2min at 9 Km/h = 18/60 km

(18/60) + (15/60) + (24/60) + (18/60) = 1.25 km = 5/4

2.5 - 1.25 = ( 10x/60 ) + (( 5 (10 - x ))/60 )

5/4 = ( 10x + 50 - 5x )/60

300/4 = 5x + 50

x = (75 - 50 )/5 = 5 km

10 min + 5 min

60*(10+5) = 900