Question

An aircraft performs a maneuver called an "aileron roll." During this maneuver, the plane turns like a screw as it maintains a straight flight path, which sets the wings in circular motion. If it takes it 35 s to complete the circle and the wingspan of the plane is 11 m, what is the acceleration of the wing tip?

Answer 1

0.17724 m/s²

Step-by-step explanation:

D = Diameter of roll = Length of wing = 11 m

T = Time it takes to complete the circle = 35 s

Velocity

`v=(2\pi R)/(T)\\\Rightarrow v=(\pi D)/(T)\\\Rightarrow v=(\pi* 11)/(35)\\\Rightarrow v=0.98735\ m/s`

Acceleration

`a=(v^2)/(R)\\\Rightarrow a=(0.98735^2)/((11)/(2))\\\Rightarrow a=0.17724\ m/s^2`

Acceleration of the tip of the plane is 0.17724 m/s²