# How much acceleration does a 747 jumbo jet of mass 6999 kg experience in takeoff when the thrust for each of four engines is 5999 N?Answer in units of m/s^2

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A plane which is taking off, or even flying, does not receive its upward thrust directly. The thrust of the engine provides a thrust which is parallel to the main axis of the plane, which at the time is parallel to the horizontal. The upward force for take-off of the plane, as well as to keep the plane in the air against the gravitational force, is provided by the force created by air acting against the wings of the speeding plane.

Thus when the plane just starts its run for the take off the entire force generated by the engines is used for accelerating the plane horizontally. If we assume that the air resistance and the rolling friction of the plane at the start of take off is zero the acceleration of the plane can be calculated as follows.

Thrust generated by 1 engine = 5999 N

Therefore total thrust generated by 4 engine = 5999*4 = 23996 N

The thrust used to accelerate the plane = (weight of plane)*acceleration =

Therefore: 6999*(acceleration) = 23996

Therefore: acceleration = 23996/6999 = 3.4285 m/s^2 (approximately)

Answer: The plane will experience acceleration of 3.4285 m/s^2

The thrust given by each of the engine is 5999N.Therefore, the total thrust by 4 engines =5999*4=23996N forward.

The graviational force on the jumbo = 6999*g N vertically down .Therefore, the frictional force on the jumbo is 6999*9.8*n back, where n is the coefficient friction of the jumbo. Therefore the resultant force -6999*9.8n+23996N .

Force = m*a, where a is the acceleration of the jumbo and m= mass of the jumbo. Given: m=6999kg, F= -6.999*9.8n+23996N and a , the acceleration is to be found out.

Therefore, 6999*a=(-6999*9.8n+23996N) or

a= (-6999*9.8n+23996)N/6999kg = (-9.8n+3.4285) m/s^2 approximately, is the acceleration of the jumbo.