Calculating Power of DC-3 Airplane Takeoff: 9500 kg Mass, 110 km/h Speed

[nm12182]

The takeoff of a DC-3 airplane is 110 km/h. Starting from rest, the ariplane takes 9.0 s to reach this speed. The mass of the (loaded) airplane is 9500 kg. What is the average power delivered by the engines to the airplane during takeoff?

Need help on this one

 

[Redbelly98]

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[thomate1]

I would approach this problem by first defining the necessary equations and variables. In this case, we can use the equation for power (P = F * v) where P is power, F is force, and v is velocity. We also need to consider the mass (m) of the airplane and the time (t) it takes to reach the given speed.

To calculate the force (F) applied by the engines, we can use the equation F = ma, where a is the acceleration of the airplane. Since we know the mass (9500 kg) and the time it takes to reach the speed (9.0 s), we can calculate the acceleration using the equation a = v/t.

Therefore, a = (110 km/h) / (9.0 s) = (110000 m/3600 s) / (9.0 s) = 3.06 m/s^2

Now, we can calculate the force applied by the engines:

F = ma = (9500 kg) * (3.06 m/s^2) = 29070 N

Finally, we can calculate the average power delivered by the engines during takeoff:

P = F * v = (29070 N) * (110000 m/3600 s) = 889.2 kW

Therefore, the average power delivered by the engines during takeoff is 889.2 kW. This is a significant amount of power and highlights the impressive capabilities of the DC-3 airplane.