A plane 3

A plane left 30 minutes later than the scheduled time. In order to reach the destination 1,500 km away on time, it had to increase its speed by 250 km/h from its usual speed. Find its usual speed.

Final Answer:

v =  750 km/h

Step-by-step explanation:

$$\begin{gathered} \Delta t=30\text{min}\to \text{h}=30:60\text{h}=0.5\text{h} \\ s=1500\text{km} \\ \Delta v=250\text{km/h} \\ s=v\cdot t=(v+\Delta v)(t-\Delta t) \\ s=v\cdot t-v\Delta t+\Delta vt-\Delta v\Delta t \\ -v\Delta t+\Delta vt-\Delta v\Delta t=0 \\ -v\Delta t+\Delta vt=\Delta v\Delta t \\ -(s/t)\Delta t+\Delta vt=\Delta v\Delta t \\ \Delta v\cdot t^2-s\cdot \Delta t=\Delta v\cdot \Delta t\cdot t \\ 250\cdot t^2-1500\cdot 0.5=250\cdot 0.5\cdot t \\ 250t^2-125t-750=0 \\ 250=2\cdot 5^3 \\ 125=5^3 \\ 750=2\cdot 3\cdot 5^3 \\ \text{GCD}(250,125,750)=5^3=125 \\ 2t^2-t-6=0 \\ a=2;b=-1;c=-6 \\ D=b^2-4ac=1^2-4\cdot 2\cdot (-6)=49 \\ D>0 \\ {t}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{1\pm \sqrt{49}}{4} \\ {t}_{1,2}=\frac{1\pm 7}{4} \\ {t}_{1,2}=0.25\pm 1.75 \\ {t}_1=2 \\ {t}_2=-1.5 \\ t=t_1=2\text{h} \\ v=s/t=1500/2=750=750\text{km/h} \\ \text{ Verifying Solution: } \\ {s}_1=v\cdot t=750\cdot 2=1500\text{km} \\ {s}_2=(v+\Delta v)\cdot (t-\Delta t)=(750+250)\cdot (2-0.5)=1500\text{km} \end{gathered}$$

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