Aircraft angines

The two engines of the aircraft are enough to supply the fuel for five hours of operation. However, one of the engines has a malfunction and thus consumes one-third more fuel. How long can the plane be in the air before it runs out of fuel? After an hour of malfunctioning, the engine will resume normal operation. How much time can a total aircraft spend in the air?

Result

t2 =  4.286 h
t3 =  4.833 h

Solution:

s=1/5/2=110=0.1 s2=s (1+1/3)=0.1 (1+1/3)2150.1333  t2=1s+s2=10.1+0.13333074.28574.286 hs=1/5 / 2=\dfrac{ 1 }{ 10 }=0.1 \ \\ s_{2}=s \cdot \ (1+1/3)=0.1 \cdot \ (1+1/3) \doteq \dfrac{ 2 }{ 15 } \doteq 0.1333 \ \\ \ \\ t_{2}=\dfrac{ 1 }{ s + s_{2} }=\dfrac{ 1 }{ 0.1 + 0.1333 } \doteq \dfrac{ 30 }{ 7 } \doteq 4.2857 \doteq 4.286 \ \text{h}
t3=1+1ss22 s=1+10.10.13332 0.12964.83334.833 ht_{3}=1 + \dfrac{ 1 - s - s_{2} }{ 2 \cdot \ s }=1 + \dfrac{ 1 - 0.1 - 0.1333 }{ 2 \cdot \ 0.1 } \doteq \dfrac{ 29 }{ 6 } \doteq 4.8333 \doteq 4.833 \ \text{h}



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