Cyclist circuit meeting

Two cyclists left one place on the circuit facing each other. The first drove at a speed of 32 km/h, and the second at 28 km/h. How long was the circuit if the cyclists met in 5 minutes?

Final Answer:

s =  5000 m

Step-by-step explanation:

$$\begin{gathered} v_1=32\text{km/h}\to \text{m/s}=32:3.6\text{m/s}=8.88889\text{m/s} \\ {v}_2=28\text{km/h}\to \text{m/s}=28:3.6\text{m/s}=7.77778\text{m/s} \\ t=5\text{min}\to \text{s}=5\cdot 60\text{s}=300\text{s} \\ s=(v_1+v_2)\cdot t=(8.8889+7.7778)\cdot 300=5000\text{m} \end{gathered}$$

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