1)The centripetal acceleration will be 84622.360 km/h².
2)The time the turn take is 0.19 hour.
3)The distance the airplane cover during the turn is 505.54 km.
What is centripetal acceleration?
The acceleration needed to move a body in a curved way is understood as centripetal acceleration.
The direction of centripetal acceleration is always in the path of the center of the course. The total acceleration is the result of tangential and centripetal acceleration.
The given data in the problem;
R is the radius= 80.5 m
v is the linear speed = 2610 km/h
a is the total acceleration.
1)
The centripetal acceleration is found by
[tex]\rm a_c = \frac{v^2}{R} \\\\ a_c = \frac{(2610 \ km/h)^2}{80.5} \\\\ a_c = 84622.360 \ km/h^2[/tex]
2)
The time the turn take is;
[tex]\rm a= \frac{4 \pi^2R}{T^2} \\\\\ 84622.360 = \frac{4 \times (3.14)^2 \times 80.5}{T^2} \\\\\ T = 0.19 \ hr[/tex]
3)
The distance the airplane cover during the turn is;
C= 2πR
C=2 × 3.14 × 80.5 km
C= 505.54 km
Hence centripetal acceleration, time the turn take, and distance the airplane cover during the turn will be 84622.360 km/h²,0.19 sec, and 505.54 km respectively.
To learn more about centripetal acceleration, refer to the link;
https://brainly.com/question/17689540
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