Solution of a)

Plane Path and Boundary Conditions

Since the velocity is tangent to the path, the rate of change of the speed is the tangential acceleration.

     at = 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

     vB = 0.75 (16)2 + 2 (16) + 176

         = 400.0 ft/s

    Solution of b)

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


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,

     an = 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.

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