Bad weather

An aircraft was slowed down during a 600 km flight due to bad weather. Its average speed was reduced by 200 km/hr from its usual speed, and the flight time increased by 30 minutes. Find the scheduled duration of the flight.

Final Answer:

t =  1 h

Step-by-step explanation:

$$\begin{gathered} s=600\text{km} \\ \Delta =200\text{km/h} \\ T=30\text{min}\to \text{h}=30:60\text{h}=0.5\text{h} \\ {v}_2=v_1-\Delta \\ {t}_2=t+T \\ t=s/v_1 \\ {t}_2=s/v_2=t+T=s/v_1+T \\ s/v_2=s/v_1+T \\ s/(v_1-\Delta )=s/v_1+T \\ s\cdot v_1=s\cdot (v_1-\Delta )+T\cdot (v_1-\Delta )\cdot v_1 \\ s\cdot x=s\cdot (x-\Delta )+T\cdot (x-\Delta )\cdot x \\ 600\cdot x=600\cdot (x-200)+0.5\cdot (x-200)\cdot x \\ -0.5x^2+100x+120000=0 \\ 0.5x^2-100x-120000=0 \\ a=0.5;b=-100;c=-120000 \\ D=b^2-4ac=100^2-4\cdot 0.5\cdot (-120000)=250000 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{100\pm \sqrt{250000}}{1}=100\pm 500\sqrt{1} \\ {x}_{1,2}=100\pm 500 \\ {x}_1=600 \\ {x}_2=-400 \\ v>0 \\ {v}_1=x_1=600\text{km/h} \\ {v}_2=v_1-\Delta =600-200=400\text{km/h} \\ t=s/v_1=600/600=1\text{h} \\ \text{ Verifying Solution: } \\ {t}_2=s/v_2=600/400=\frac{3}{2}=1.5\text{h} \end{gathered}$$

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