Sawmill factory

Peter works in the factory. The bus stop is 10 km from the factory. Therefore, when the bus arrives for Peter, the driver always leaves the factory and takes him to work. They arrived at the saw exactly at 8:00. Today, the bus arrived 11 minutes earlier, and Petr walked to the meeting car from the sawmill. His walk is 10x slower than the speed of the car. At what time did Peter arrive at the sawmill?

Result

T = 7:56

Step-by-step explanation:

$$\begin{gathered} v_1=55\text{km/h} \\ {v}_2=v_1/10=55/10=\frac{11}{2}=5\frac{1}{2}=5.5\text{km/h} \\ a=11/60=\frac{11}{60}\doteq 0.1833\text{h} \\ s=10\text{km} \\ {t}_1=s/v_1=10/55=\frac{2}{11}\doteq 0.1818\text{h} \\ {T}_1=8.00-t_1=8.00-0.1818=\frac{86}{11}=7\frac{9}{11}\doteq 7.8182\text{h} \\ {s}_1+s_2=s \\ {v}_1{t}_2+v_2\cdot (t_2+a)=s \\ {t}_2=(s-v_2\cdot a)/(v_1+v_2)=(10-5.5\cdot 0.1833)/(55+5.5)\doteq 0.1486 \\ {s}_1=t_2\cdot v_1=0.1486\cdot 55=\frac{1079}{132}=8\frac{23}{132}\doteq 8.1742 \\ {s}_2=(t_2+a)\cdot v_2=(0.1486+0.1833)\cdot 5.5=\frac{241}{132}=1\frac{109}{132}\doteq 1.8258 \\ T=T_1-a+2\cdot t_2=7.8182-0.1833+2\cdot 0.1486=7.93209=7:56 \end{gathered}$$

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