Cross road

From the junction of two streets that are perpendicular to each other, two cyclists (each on another street) walked out. One ran 18 km/h and the second 24 km/h. How are they away from

a) 6 minutes,
b) 15 minutes?

Result

a =  3 km
b =  7.5 km

Solution:

$v_{ 1 } = 18 \ km/h \ \\ v_{ 2 } = 24 \ km/h \ \\ \ \\ v = \sqrt{ v_{ 1 }^2 + v_{ 2 }^2 } = \sqrt{ 18^2 + 24^2 } = 30 \ km/h \ \\ t_{ 1 } = 6 \ min = 6 / 60 \ h = 0.1 \ h \ \\ a = t_{ 1 } \cdot \ v = 0.1 \cdot \ 30 = 3 = 3 \ \text{ km }$

$t_{ 2 } = 15 \ min = 15 / 60 \ h = 0.25 \ h \ \\ \ \\ b = t_{ 2 } \cdot \ v = 0.25 \cdot \ 30 = \dfrac{ 15 }{ 2 } = 7.5 = 7.5 \ \text{ km }$

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