Answer:
a) a = 2.096 m/s² = 2.1 m/s²
b) Take off time = 15.9 s
Explanation:
Using the equations of motion
x = take off run = 265 m
Initial velocity, u = 0 m/s (plane starts from rest)
final velocity = lift off speed = v = 120 km/h = 33.33 m/s
a = ?
a) v² = u² + 2ax
33.33² = 0² + 2×a×265
a = 1111.1111/530
a = 2.096 m/s²
b) v = u + at
33.33 = 0 + (2.096 × t)
t = 33.33/2.096
t = 15.9 s
