Question

A small aircraft accelerated down a runway at 4.0 m/s²

Answer 1

Given data in the problem :-

• Acceleration (a) = 4.0 m/s^2
• Initial velocity (u) = 0 m/s
• Final velocity (v) = 34 m/s
• Distance travelled by aircraft (S) =  ?

From Newton's Laws of Motion we know that ,

v = u + at  [t = Time taken by aircraft to cover the distance]

⇒ 34 = 0 + 4t

⇒ t = 34/4 s

∴  t = 8.5 s

From Newton's Laws of Motion we also know that ,

S = u.t + 1/2a.t^2

⇒ S = 0×8.5 + 1/2 × 4 × (8.5)^2 m

∴  S = 144.50 m

Thus the distance travelled by the aircraft while accelerating is 144.50 meter .