Much-expected 9871

The train travels at an average speed of 75 km/h. According to the timetable, he should be at the station in 11 minutes, but he still has 20 km to go. How much-expected delay will appear on the station information board?

Correct answer:

d =  5 min

Step-by-step explanation:

$$\begin{gathered} v=75\text{km/h} \\ {t}_1=11\text{min}\to \text{h}=11:60\text{h}=0.18333\text{h} \\ s=20\text{km} \\ {t}_2=s/v=20/75=\frac{4}{15}\doteq 0.2667\text{h} \\ {d}_1=t_2-t_1=0.2667-0.1833=\frac{1}{12}\doteq 0.0833\text{h} \\ d=d_1\to \text{min}=d_1\cdot 60\text{min}=0.0833\cdot 60\text{min}=5\text{min}=5\text{min} \end{gathered}$$

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