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Final Exam Review: Motion of an Airplane |
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| May5-12, 11:11 PM | #1 |
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Final Exam Review: Motion of an Airplane
This is a question from an old test that I took this semester. My grader did not input anything, so I am left to trying to figure out by myself if my answers are correct or not.
Question: The motion of an airplane leaving from miami can be described using a cartesian coordinate system with due East coincendet with the positive x-direction and due Noth with the positrive y-direction. The position vector r of the airplane as a function of time from take-off can then be written as: vector r = at(i) + (Bt - yt^3)(j) [a.] What are the units of the three constants? This one was relatively simple. I found that the units were: a = m/s. B = m/s. and y = m/s^3. [b.] Find the time(s) when the motion of the airplane is due NE. By motion I am guessing they mean velocity, So Vx = a and Vy = B - 3yt^2. Because it is exactly NE, these two components should be equal to one another, giving Vx = Vy ∴ a = B - 3yt^2, t = [(a - B)/3y]^1/2. The negative answer for t can be discarded. [c.] Find the plane's position when the motion is due NE. Position means the angle. So theta = tan^-1(ry/rx) Which is tan^-1([Bt - yt^3]/at) ----------------------------------------------------------------------------------------- Please help me confirm whether I am wrong or right as my final is on Monday and I need to do extremely well in order to pass this course Thanks. |
| May6-12, 02:26 AM | #2 |
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