Twelve-minute run

Runner Adam performs a twelve-minute training run every day. He starts running at a speed of 3.5 m/s and every three minutes he increases his speed by 0.5 m/s. At the end of the first section, the cyclist Eva overtakes the runner, who is riding at a constant speed of 4.2 m/s.
How long are the individual training sections?
How many metres does Adam run in total? At what distance from the starting place will Adam catch up with Eva?
What distance will there be between Adam and Eva when the runner finishes his run?
What distance was there between Adam and Eva when the runner began his run? Eva was already on her bicycle at that time.

Final Answer:

s₁ =  2835 m
o =  3060 m

Step-by-step explanation:

$$\begin{gathered} v_1=3.5\text{m/s} \\ {t}_1=3\text{min}\to \text{s}=3\cdot 60\text{s}=180\text{s} \\ e=4.2\text{m/s} \\ a=v_1\cdot t_1=3.5\cdot 180=630\text{m} \\ {v}_2=v_1+0.5=3.5+0.5=4\text{m/s} \\ b=v_2\cdot t_1=4\cdot 180=720\text{m} \\ {v}_3=v_2+0.5=4+0.5=\frac{9}{2}=4.5\text{m/s} \\ c=v_3\cdot t_1=4.5\cdot 180=810\text{m} \\ {v}_4=v_3+0.5=4.5+0.5=5\text{m/s} \\ {s}_1=s_2 \\ a+b+c+v_4\cdot t_4=e\cdot (t_1+t_1+t_1+t_4) \\ 630+720+810+5\cdot t_4=4.2\cdot (180+180+180+t_4) \\ 0.8t_4=108 \\ {t}_4=\frac{108}{0.8}=135 \\ {t}_4=135 \\ {t}_4<t_1 \\ {s}_1=a+b+c+v_4\cdot t_4=630+720+810+5\cdot 135=2835\text{m} \end{gathered}$$

$$\begin{gathered} d=v_4\cdot t_1=5\cdot 180=900\text{m} \\ o=a+b+c+d=630+720+810+900=3060\text{m} \end{gathered}$$

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