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Final Exam Review: Motion of an Airplane

by Metamorphose
Tags: airplane, exam, final, motion, review
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May5-12, 11:11 PM
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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.


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

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May6-12, 02:26 AM
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Quote Quote by Metamorphose View Post
[c.] Find the plane's position when the motion is due NE.
Position means the angle.
I would have thought position meant position in 2 dimensions, probably in Cartesian co-ordinates.

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