Cyclists and walkers

A group of tourists started at 8:00 at speed 4 km/h walk. At half-past ten, another group started on a bike and caught up with a group of tourists at 10:50. What was the average speed of cyclists?

Correct result:

v₂ =  8.5 km/h

Solution:

$$\begin{gathered} v_1=4\text{ km/h } \\ {t}_3=10+50\mathrm{/}60={\displaystyle \frac{65}{6}}\doteq 10.8333\text{ h } \\ {t}_1=t_3-8.00=10.8333-8.00={\displaystyle \frac{17}{6}}\doteq 2.8333\text{ h } \\ {t}_2=t_3-(9+30\mathrm{/}60)=10.8333-(9+30\mathrm{/}60)={\displaystyle \frac{4}{3}}\doteq 1.3333\text{ h } \\ {s}_1=s_2 \\ {v}_1\cdot t_1=v_2\cdot t_2 \\ {v}_2=v_1\cdot {\displaystyle \frac{t_1}{t_2}}=4\cdot {\displaystyle \frac{2.8333}{1.3333}}={\displaystyle \frac{17}{2}}={\displaystyle \frac{17}{2}}\text{ km/h}=8.5\text{ km/h} \end{gathered}$$

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