Sawmill factory

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


T = 7:56 Wrong answer

Step-by-step explanation:

v1=55 km/h v2=v1/10=55/10=211=521=5.5 km/h a=11/60=60110.1833 h s=10 km t1=s/v1=10/55=1120.1818 h T1=8.00t1=8.000.1818=1186=71197.8182 h  s1+s2 = s v1 t2 + v2   (t2 + a) = s t2=(sv2 a)/(v1+v2)=(105.5 0.1833)/(55+5.5)0.1486 s1=t2 v1=0.1486 55=1321079=8132238.1742 s2=(t2+a) v2=(0.1486+0.1833) 5.5=132241=11321091.8258  T=T1a+2 t2=7.81820.1833+2 0.1486=7.93209=7:56

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