Cross road

From the junction of two streets perpendicular to each other, two cyclists (each on another street) walked out. One ran at 18 km/h, and the second at 24 km/h. How are they away from a) 6 minutes, b) 15 minutes?

Final Answer:

a =  3 km
b =  7.5 km

Step-by-step explanation:

$$\begin{gathered} v_1=18\text{km/h} \\ {v}_2=24\text{km/h} \\ {v}^2=v_1^2+v_2^2 \\ v=\sqrt{v_1^2+v_2^2}=\sqrt{18^2+24^2}=30\text{km/h} \\ {t}_1=6\text{min}\to \text{h}=6:60\text{h}=0.1\text{h} \\ a=t_1\cdot v=0.1\cdot 30=3\text{km} \end{gathered}$$

$$\begin{gathered} t_2=15\text{min}\to \text{h}=15:60\text{h}=0.25\text{h} \\ b=t_2\cdot v=0.25\cdot 30=7.5\text{km} \end{gathered}$$

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