Answer:
2.3125m/s²
Explanation:
Using the equation of motion v² = u²+2aS
v is the final velocity = 120km/hr
120km/hr = 120 * 1000/1 * 3600 = 33.3m/s
u is the initial velocity = 0m/s
a is the acceleration
S is the distance covered = 240m
On substituting the given parameters
33.3² = 0²+2a(240)
33.3² = 480a
1110 = 480a
a = 1110/480
a = 2.3125m/s²
Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²
