Respuesta :
Answer:
The magnitude of the centripetal acceleration during the turn is [tex]a=12.04\ m/s^2.[/tex]
Explanation:
Given :
Speed to the airplane in circular path , v = 115 m/s.
Time taken by plane to turn , t= 15 s.
Also , the plane turns from east to south i.e. quarter of a circle .
Therefore, time taken to complete whole circle is , [tex]T=t\times 4=60\ s.[/tex]
Now , Velocity ,
[tex]v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.[/tex]
Also , we know :
Centripetal acceleration ,
[tex]a=\dfrac{v^2}{r}[/tex]
Putting all values we get :
[tex]a=12.04\ m/s^2.[/tex]
Hence , this is the required solution .
