Plane Path and Boundary Conditions


Since the velocity is tangent to the path, the rate of change of the
speed is the tangential acceleration.
a_{t} = dv/dt = 2 + 1.5t ft/s^{2}
The tangential acceleration can be integrated to obtain an equation for the speed as a function of time. Since the plane enters the turn traveling at 176 ft/s, this becomes the lower limit of integration for the velocity,
v = ds/dt = 0.75 t^{2} + 2 t +
176
Substituting the time required for the plane to complete the turn, the speed at B becomes
v_{B} = 0.75 (16)^{2} + 2 (16) + 176
= 400.0 ft/s

Normal Acceleration of Plane


The velocity equation above can be integrated to obtain an equation for the displacement s as a function of time.
s = 0.25 t^{3} + t^{2}
+ 176 t
Again substituting the time required, the distance traveled
during the turn is obtained as
s = 0.25 (16)^{3} + (16)^{2}
+ 176 (16)
= 4,096 ft
Since the plane travels in a semicircular path and the total distance traveled is now known, the radius of of the path can be determined as
2 π R = 2 (4,096
ft)
R = 1,304
ft
The radius of curvature is a constant equal to the radius of the semicircular flight path:
ρ = R = 1,304 ft

Acceleration in the Normal Direction
over time (expresses as g's) 

The maximum centripetal acceleration occurs at point B, where the
speed v is a maximum,
a_{n} = v^{2}/ρ = v^{2}/R
= 400^{2}/1,304 = 122.7 ft/s^{2}
The acceleration of gravity on Earth is constant at 32.2 ft/s^{2}. At 122.7 ft/s^{2}, the maximum centripetal acceleration that the plane experiences is approximately 4 times that due to gravity. It is unlikely that a cargo plane would execute such a "highg" turn. 