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During a takeoff run, an aircraft starts from rest and has a lift-off speed of 120 km/h. a. What minimum constant acceleration does the aircraft require if the aircraft is to be airborne after a takeoff run of 280 m?

User GoGoris
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1 Answer

5 votes

Answer:

1.984 m/s^2

Step-by-step explanation:

initial velocity of air craft, u = 0 m/s

final speed of the aircraft, v = 120 km/h

Convert the speed into m/s from km/h

So, v = 120 km/h = 33.33 m/s

distance, s = 280 m

Let a be the acceleration of the aircraft.

Use third equation of motion


v^(2)=u^(2)+2* a* s


33.33^(2)=0^(2)+2* a* 280

a = 1.984 m/s^2

Thus, the acceleration of the aircraft is 1.984 m/s^2.

User Majid Shahabfar
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