What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of 265 m?

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

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s