A DC-9 performs a manueber over a 65 second duration that allows for 25 seconds
ID: 2126259 • Letter: A
Question
A DC-9 performs a manueber over a 65 second duration that allows for 25 seconds of "weightlessness" during a central segment. A set of parametic equations describe the planes position vectore during this 25 second segment as:
<x,y> = <122.63t, 7315.2 +122.63t - 4.905t^2>
Determine the vector direction and magnitude of the radius of curvature of this path at t=5 seconds and t=20 seconds. Also calculate the vecot acceleration at all times during the "weightlessness" segment. Why do people experience "weightlessness" during this maneuver?
Explanation / Answer
<x,y> = <122.63t, 7315.2 +122.63t - 4.905t^2>
(Vx , Vy) = (122.63 , 122.63 - 9.81t>
At t= 5
V = <122.63 , 122.63-9.81x5) = < 122.63 , 73.58>
At t = 20
V = <122.63 . 122.63-9.81x20> = < 122.63 ,- 73.58 >
Acceleration vector = < 0 , -9.81 >
Since acceleration is negative , thy feel weightlessness
Radius of carvature
|| v x a|| /(||a||^3 = || (122.63 , 122.63 - 9.81t> x < 0 , -9.81 > || /(9.81||^3
= || -122.63 x 9.81 || || /(9.81||^3
= 122.63/9.81^2 = 1.274
At t = 5 , Radius of carvature = 1.274
At t= 20 , radius of carvature = 1.274
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