Power during takeoff

The aircraft weighing 3.5 tons will disembark 1 km in 1 minute after takeoff and reach a speed of 290 km/h. Find the average power of its engines during this time.

Correct answer:

P =  761.52 kW

Step-by-step explanation:

$$\begin{gathered} m=3.5\text{t}\to \text{kg}=3.5\cdot 1000\text{kg}=3500\text{kg} \\ h=1\text{km}\to \text{m}=1\cdot 1000\text{m}=1000\text{m} \\ v=290\text{km/h}\to \text{m/s}=290:3.6\text{m/s}=80.55556\text{m/s} \\ t=1\text{min}\to \text{s}=1\cdot 60\text{s}=60\text{s} \\ g=9.81{\text{m/s}}^2 \\ {E}_1=m\cdot g\cdot h=3500\cdot 9.81\cdot 1000=34335000\text{J} \\ {E}_2=\frac{1}{2}\cdot m\cdot v^2=\frac{1}{2}\cdot 3500\cdot {\frac{725}{9}}^2\doteq 11356095.679\text{J} \\ W=E_1+E_2=34335000+\frac{283902392}{25}=\frac{34335000\cdot 25}{25}+\frac{283902392}{25}=\frac{858375000}{25}+\frac{283902392}{25}=\frac{858375000+283902392}{25}=\frac{1142277392}{25}\doteq 45691095.679\text{J} \\ {P}_1=W/t=\frac{1142277392}{25}/60=\frac{1142277392}{25\cdot 60}=\frac{1142277392}{1500}\doteq 761518.2613\text{W} \\ P=P_1\to \text{kW}=P_1:1000\text{kW}=761518.2613:1000\text{kW}=761.52\text{kW}=761.52\text{kW} \end{gathered}$$

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