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?

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=8.5  km/h v_{ 1 } = 4 \ km/h \ \\ t_{ 3 } = 10+50/60 = \dfrac{ 65 }{ 6 } \doteq 10.8333 \ h \ \\ t_{ 1 } = t_{ 3 } - 8.00 = 10.8333 - 8.00 = \dfrac{ 17 }{ 6 } \doteq 2.8333 \ h \ \\ t_{ 2 } = t_{ 3 } - (9+30/60) = 10.8333 - (9+30/60) = \dfrac{ 4 }{ 3 } \doteq 1.3333 \ 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 = 8.5 \ \text { km/h }

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