Question

A small turbo-prop commuter airplane, starting from rest on a Lansing airport runway, accelerates for 22.5s before taking off. Its speed at takeoff is 53.0 m/s (119 mi/hr). Calculate the acceleration of the plane, in g's, assuming it remains constant. (i.e., divide the acceleration in m/s2 by 9.81 m/s2).In the problem above, how far did the plane move while accelerating for 22.5 s?

Answer 1

s = 596.25 m

Step-by-step explanation:

given,

initial speed, u = 0 m/s

final speed, v = 53 m/s

time, t = 22.5 s

acceleration of the plane = ?

`a = (v-u)/(t)`

`a = (53-0)/(22.5)`

a = 2.36 m/s²

using equation of motion

`s = u t + (1)/(2)at^2`

`s =(1)/(2)* 2.36* 22.5^2`

s = 596.25 m

Hence, the distance traveled by the plane is equal to 596.25 m