# 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:

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

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