Because the takeoff roll is a straight line, this problem can be analyzed using rectangular coordinates.
Begin with a freebody diagram. Let the origin of the rectangular coordinate system be located at the point where the plane begins the takeoff roll, with the positive x axis in the direction of the takeoff roll. Assume there is no force or velocity in the z direction.
At t = 0, the velocity of the F16 is 0,
(v_{x})_{t=0} =
(v_{y})_{t=0} = 0
Use the value of the thrust at t = 0 and t = 5 to express the thrust in the xdirection during the first 5 seconds as a function of time, giving
T_{t=0} = 22,250i
N
T_{t=5} = 111,250i
N
T(t) = (22,250 + 17,800t)i N (0 ≤
t ≤ 5 s)
The velocity in the y direction is zero before take off and does not effect the velocity in the x direction. Likewise, forces in the y direction are independent of the thrust before take off.
The change in momentum in the x direction is equal to the thrust impulse. This relationship can be used to determine the horizontal velocity for first 5 seconds,
The total velocity after the first 5 seconds of the takeoff roll is
v_{t=5} = 22.86i m/s
