# Movement

From the crossing of two perpendicular roads started two cyclists (each at different road). One runs at average speed 28 km/h, the second at average speed 24 km/h. Determine the distance between them after 45 minutes cycling.

Result

x =  27.659 km

#### Solution:

$t = 45 \ min = 45 / 60 \ h = 0.75 \ h \ \\ v_{ 1 } = 28 \ km/h \ \\ v_{ 2 } = 24 \ km/h \ \\ \ \\ s_{ 1 } = v_{ 1 } \cdot \ t = 28 \cdot \ 0.75 = 21 \ km \ \\ s_{ 2 } = v_{ 2 } \cdot \ t = 24 \cdot \ 0.75 = 18 \ km \ \\ \ \\ x = \sqrt{ s_{ 1 }^2 + s_{ 2 }^2 } = \sqrt{ 21^2 + 18^2 } = 3 \ \sqrt{ 85 } \doteq 27.6586 = 27.659 \ \text { km }$

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