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/s²

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 t² + 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 t³ + t² + 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

The maximum centripetal acceleration occurs at point B, where the speed v is a maximum,

     a_n = v²/ρ = v²/R

         = 400²/1,304 = 122.7 ft/s²

The acceleration of gravity on Earth is constant at 32.2 ft/s². At 122.7 ft/s², 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 "high-g" turn.