Two runners

Two runners ran simultaneously towards each other from locations distant 21.4 km. The average speed of the first runner was 1/6 higher than the average speed of the second runner.

How long should each run a 21.4 km, if you know they meet after 78 minutes?

Correct answer:

t₁ =  144.9 min
t₂ =  169 min

Step-by-step explanation:

$$\begin{gathered} t=78\text{min} \\ s=21.4\text{km} \\ v=s/t=21.4/78=\frac{107}{390}\doteq 0.2744\text{km/min} \\ {s}_1+s_2=s \\ {v}_1=v_2(1+1/6)=1.1667v_2 \\ {v}_1=1.1667\cdot v_2 \\ {v}_1+v_2=v \\ {v}_1=1.1667\cdot v_2 \\ {v}_1+v_2=0.27435897435897 \\ {v}_1-1.1667v_2=0 \\ {v}_1+v_2=0.274359 \\ Row2-Row1\to Row2 \\ {v}_1-1.17v_2=0 \\ 2.17v_2=0.27 \\ {v}_2=\frac{0.27435897}{2.1667}=0.12662527 \\ {v}_1=0+1.1667v_2=0+1.1667\cdot 0.12662527=0.1477337 \\ {v}_1=0.147734 \\ {v}_2=0.126625 \\ {t}_1=t_1=s/v_1=21.4/0.1477=144.9\text{min} \end{gathered}$$

$t_2=t_2=s/v_2=21.4/0.1266=169\text{min}$

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