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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

a million. 1800=(a million/2)(5m/s^2)t = 26.8s 2. v = (5m/s^2)(26.8s) = 134m/s 3. d = (a million/2)(5m/s^2)(1s)^2 = 2.5m 4. d = (a million/2)(5m/s^2)(25.8s)^2 = 132m 5. what number of the takeoff speed did the airplane benefit whilst it reached the midpoint of the runway? 1800/2 = 900 v^2 = u^2 + 2a(delta x) 5^2 = u^2 + 2(5m/s^2)(900m) u^2 = ninety 4.74m/s ninety 4.seventy 4/134 = 70.7%