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?

Correct result:

a =  3 km
b =  7.5 km

#### Solution:

$v_{1}=18 \ \text{km/h} \ \\ v_{2}=24 \ \text{km/h} \ \\ \ \\ v=\sqrt{ v_{1}^2 + v_{2}^2 }=\sqrt{ 18^2 + 24^2 }=30 \ \text{km/h} \ \\ t_{1}=6 \ min \rightarrow h=6 / 60 \ h=0.1 \ h \ \\ a=t_{1} \cdot \ v=0.1 \cdot \ 30=3 \ \text{km}$
$t_{2}=15 \ min \rightarrow h=15 / 60 \ h=0.25 \ h \ \\ \ \\ b=t_{2} \cdot \ v=0.25 \cdot \ 30=\dfrac{ 15 }{ 2 }=7.5 \ \text{km}$

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