If the same airplane as above airplane starts from a brakes locked position on the runway and it takes off at an airspeed of 200fps. what is the time for the aircraft to reach take-off speed?

t = 16.7s

Using the the acceleration for the airplane as above, that is

a = 12 ft/s²

Now,

let t = the time it takes for the airplane to reach that take off speed

Vo = 0 ft/s (see explanation below)

V = 200ft/s

V = Vo + at (see explanation below)

t = (200ft/s - 0 ft/s) / 12ft/s²

t = 200ft/s / 12ft/s²

t = 16.7 seconds

Explanation:

The solution to this problem assumes that the force propelling the airplane forward is constant, and as a result the acceleration too must be constant.

If the acceleration of the airplane is constant, then the equations of constant acceleration motion applies here so V = Vo + at

The initial speed, Vo, is equal to zero because the airplane was in a brakes locked position and so it starts its motion before take-off from rest.

Thank you for reading and I hope this was helpful to you.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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