Same route

The tourist came out at 6:00 at an average speed of 4 km/h. In 2 hours, a cyclist set off on the same route at an average of 28 km/h. When did the cyclist catch the tourist?

Result

t = 8:20 hh:mm

Step-by-step explanation:

$$\begin{gathered} v_1=4\text{km/h} \\ {v}_2=28\text{km/h} \\ {s}_1=s_2 \\ {v}_1\cdot t_1=v_2\cdot t_2 \\ {v}_1\cdot (t-6)=v_2\cdot (t-(6+2)) \\ 4\cdot (t-6)=28\cdot (t-(6+2)) \\ 24t=200 \\ t=\frac{200}{24}=8.33333333 \\ t=\frac{25}{3}\doteq 8.333333=8.33333=8:20\text{hh}:\text{mm} \end{gathered}$$

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