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?

Correct result:

a =  3 km
b =  7.5 km

Solution:

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

$$\begin{gathered} t_2=15\text{ min}\to \text{ h}=15\mathrm{/}60\text{ h}=0.25\text{ h } \\ b=t_2\cdot v=0.25\cdot 30={\displaystyle \frac{15}{2}}={\displaystyle \frac{15}{2}}\text{ km}=7.5\text{ km} \end{gathered}$$

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