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I know the answer of a; how do I find the equation for the distance between the planes at all times for part b?
Is it solvable?
its solveble, buti have to read thru it first, and these old eyes get tired at times
both at 33000 ft they intersect directly over Frada hgts near misses shortly after 130pm --------------------------------------- at 130, Am2003 was 32 Nm from Frada, heading 171o at a rate of 405knots at 130 Un366 was 44 Nm from Frada, heading 81o at a rate of 465 knots
how how do we read airplane headings? from the north line right?
doesnt really matter, the orientation is the same regardless of our 'proper' placements
lets draw a picture, both planes, and Frada
Ohh ok I just assumed it was a right triangle and so drew a right triangle with 81 9 and 90
171-81 = 90, so your assumption was valid
we have that, we got the rate of the distance decreasing between them two, but we need to find the actual distance separation at the closest point. that is the part we are having trouble with
the rate we got is 472.8 knots
yup we have that already
d^2 = x^2 + y^2 rate of change is derivatives, so what is our derivative?
jut making sure we are on the same page here
the shortest distance between them is when one of them is at Frada ... that is when they are in line with each other.
what is our derivative? lets make sure we are working it thru and have not developed any errors
hmm..how did you draw the triangle?
yup we have that
we got part a already, we got ds/dt = 494.7
yyup thanks we already have that part; we got ds/dt = 497.7 already. we are only stuck on part b - thank you so much!
for part b) i set AA = 32-405t distance from Frada UA = 44 -465t distance from frada Distance between AA and UA D= sqrt((32-405*t)^2+(44-465*t)^2) minimize this by taking the derivative dD/dt = (1/2)*(-66840+760500*t)/sqrt((32-405*t)^2+(44-465*t)^2) set dD/dt = 0
that makes sense! we got t = .0879
@amistre64 please check for part a) i got a different ds/dt x^2 + y^2 = s^2 2x x' + 2y y' = 2s s' x* x' + y* y' = s * s ' 44 * (-465) + 32 * (-405) = sqrt( 44^2 + 32^2 ) * s ' s' = [44 * (-465) + 32 * (-405)]/ sqrt( 44^2 + 32^2 ) s' = -614.27 knots
right, t = 1114/12675 = .0878 is the time when their distance is closest. and if you plug in you get D = 4.7677 miles .
do you agree with my work, how i got ds/dt = -614.27
ohh!! we screwed it up thank you!!
44 * (-465) + 32 * (-405) I think you had an error here...
I think the speed of American 1003 is 405 not 465
we can think of this problem as two dots approaching each other. The path of the dots can be represented by parametric equations
woops i need a negative sign there
do you know "whether ATC would have time to take action? Might a slight altitude change for one of the lights help prevent any five mile rule violation?"
t=0 is 1:30 PM
+465 since it is moving to the right
now if you plug in t= .08 into both equations you can see exactly where the points are
clearly AA gets to Frada before UA because 32-405t = 0 , t= 32/405 ~ .079 sec -44 + 465t = 0, t= 44/464 ~ .094 sec but at t = 1114/12675 they are closest together plug in this time into AA and UA AA is below F by 3.59 nautical miles UA is to the left of Frada by 3.13 nautical miles
if you change the altitude that could change the minimum 5 nautical miles rule. but we would have to go into three dimensions
without loss of generality we can set one of the planes to 30,000 feet and the other one to some value A. But note that we must change units to nautical miles to be consistent with our earlier calculations.