What is the distance traveled by the plane in the last second before taking off?

I have V takeoff=134 m/s, t takeoff=26.83,


A plane accelerates from rest at a constant rate of 5.00 meters/ seconds squared along a runway that is 1800M long. Assume that the plane reaches the required takeoff velocity at the end of the runway.|||131.5 m. To get the distance you can first find the velocity v0 at 1 sec before takeoff from acceleration a = 5 m/s^2, time window t= 1 s, and final velocity v1 = 134 m/s. Then find distance s from average velocity * time.


v0 = v1 - at = 129 m/s


distance s = (v0 + v1)/2 * t = 131.5 m|||distance traveled by the plane in the last second before taking off = 1800m - distance traveled by the plane in 25.83





distance traveled by the plane in 25.83


d = vinitial t + 0.5at^2


d = 0 + 0.5(5)(25.83^2)


d = 1667.97 m





distance traveled by the plane in the last second before taking off = 1800 - 1667.97 = 132 m