The speed of the plane shown below is 500 m/s. The rate of change of the path an
ID: 1828548 • Letter: T
Question
The speed of the plane shown below is 500 m/s. The rate of change of the path angle (theta) is constant and equal to 5 deg/s.
a. Calculate the tangential and normal components of the acceleration of this plane.
b. Calculate the radius of curvature of the plane at this instant in time.
The speed of the plane shown below is 500 m/s. The rate of change of the path angle (theta) is constant and equal to 5 deg/s. Calculate the tangential and normal components of the acceleration of this plane. Calculate the radius of curvature of the plane at this instant in time.Explanation / Answer
w = dtheta/dt = 5 deg/s = 0.0873 rad/s
v = 500 m/s
v = wr
r = radus of curvature
500 = 0.0873 x r
r = 5729.577 m
normal acc. = v^2/r = 500x 500 / 5729.577 = 43.633 m/s^2
tangential acceleration
as dtheta/dt = constant i.e w = constant thus angular accelration = 0
tnagential acc. = alpha x r
as alpha = 0
tangential acc. = 0
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