Two runners

Two runners ran simultaneously towards each other from locations distant 34.6 km. The average speed of the first runner was 1/5 higher than the average speed of the second runner.

How long should each ran a 34.6 km, if you know that they meet after 67 minutes?

Correct result:

t₁ =  122.8 min
t₂ =  147.4 min

Solution:

$$\begin{gathered} t=67min \\ {s}_1+s_2=34.6km \\ {v}_1=v_2(1+1\mathrm{/}5)=1.2v_2 \\ {v}_{1}t+v_{2}t=34.6 \\ {v}_1+v_2=34.6\mathrm{/}67=0.5164km\mathrm{/}min \\ 2.2v_2=0.5164 \\ {v}_2=0.2347km\mathrm{/}min \\ {v}_1=1.2v_2=0.2817km\mathrm{/}min \\ {t}_1=s\mathrm{/}{v}_1=34.6\mathrm{/}0.2817=122.8\text{ min} \end{gathered}$$

$t_2=s\mathrm{/}{v}_2=34.6\mathrm{/}0.2347=147.4\text{ min}$

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