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 v1=28 km/h v2=24 km/h  s1=v1 t=28 0.75=21 km s2=v2 t=24 0.75=18 km  x=s12+s22=212+1823 8527.658627.659 kmt=45 \ min=45 / 60 \ h=0.75 \ h \ \\ v_{1}=28 \ \text{km/h} \ \\ v_{2}=24 \ \text{km/h} \ \\ \ \\ s_{1}=v_{1} \cdot \ t=28 \cdot \ 0.75=21 \ \text{km} \ \\ s_{2}=v_{2} \cdot \ t=24 \cdot \ 0.75=18 \ \text{km} \ \\ \ \\ x=\sqrt{ s_{1}^2 + s_{2}^2 }=\sqrt{ 21^2 + 18^2 } \doteq 3 \ \sqrt{ 85 } \doteq 27.6586 \doteq 27.659 \ \text{km}



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