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:

v1=4 km/h t3=10+50/60=656≐10.8333 h t1=t3−8.00=10.8333−8.00≐176≐2.8333 h t2=t3−(9+30/60)=10.8333−(9+30/60)≐43≐1.3333 h  s1=s2 v1⋅ t1=v2⋅ t2  v2=v1⋅ t1t2=4⋅ 2.83331.3333=172=8.5 km/hv_{1}=4 \ \text{km/h} \ \\ t_{3}=10+50/60=\dfrac{ 65 }{ 6 } \doteq 10.8333 \ \text{h} \ \\ t_{1}=t_{3} - 8.00=10.8333 - 8.00 \doteq \dfrac{ 17 }{ 6 } \doteq 2.8333 \ \text{h} \ \\ t_{2}=t_{3} - (9+30/60)=10.8333 - (9+30/60) \doteq \dfrac{ 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 \ \dfrac{ t_{1} }{ t_{2} }=4 \cdot \ \dfrac{ 2.8333 }{ 1.3333 }=\dfrac{ 17 }{ 2 }=8.5 \ \text{km/h}

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