Two runners

Two runners set off simultaneously toward each other from locations 23.1 km apart. The average speed of the first runner was 1/7 greater than the average speed of the second runner.

How long would each take to run 23.1 km, given that they meet after 58 minutes?

Final Answer:

t₁ =  108.7 min
t₂ =  124.3 min

Step-by-step explanation:

$$\begin{gathered} t=58\text{min} \\ s=23.1\text{km} \\ v=s/t=23.1/58=\frac{231}{580}\doteq 0.3983\text{km/min} \\ k=1+\frac{1}{7}=\frac{1\cdot 7}{7}+\frac{1}{7}=\frac{7}{7}+\frac{1}{7}=\frac{7+1}{7}=\frac{8}{7}\doteq 1.1429 \\ {s}_1+s_2=s \\ {v}_1=v_2(1+1/7)=k{v}_2 \\ {v}_1=k\cdot v_2 \\ {v}_1+v_2=v \\ k\cdot v_2+v_2=v \\ 1.1428571428571\cdot v_2+v_2=0.39827586206897 \\ 2.142857v_2=0.398276 \\ {v}_2=\frac{0.39827586}{2.14285714}=0.18586207 \\ {v}_2=\frac{539}{2900}\doteq 0.185862 \\ {v}_1=v-v_2=0.3983-0.1859=\frac{154}{725}\doteq 0.2124\text{km/min} \\ {t}_1=t_1=s/v_1=23.1/0.2124=108.7\text{min} \end{gathered}$$

$t_2=t_2=s/v_2=23.1/0.1859=124.3\text{min}$

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