Peleton

The cycling race was run at an average speed of 42 km/h. One cyclist lost due to a defect 9 minutes. How long and far must he goes at speed 47 km/h to catch the peloton again?

Final Answer:

t =  75.6 min
s =  59.22 km

Step-by-step explanation:

$$\begin{gathered} v_1=42\text{km/h} \\ {v}_2=47\text{km/h} \\ d=9\text{min} \\ {s}_1=s_2=s \\ {v}_1(t+d)=v_{2}t \\ {v}_{1}t+d{v}_1=v_{2}t \\ t=d\cdot \frac{v_1}{v_2-v_1}=9\cdot \frac{42}{47-42}=75.6\text{min} \end{gathered}$$

$$\begin{gathered} t_2=t\to \text{h}=t:60\text{h}=75.6:60\text{h}=1.26\text{h} \\ s=v_2\cdot t_2=47\cdot 1.26=59.22\text{km} \end{gathered}$$

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