"At time 1050 you see another aircraft on a bearing of 040° at a range of 10 miles. If at 1056 the bearing remains 040° but the range has reduced to 8 miles. At what time will the range reduce to 1 mile, assuming the bearing remains constant?" Got 1117 hrs.
sounds like a calculus problem.... you able to do calculus and related rates yet?
otherwise, we can do right triangles to figure speed and such
ermm.. No. I have only done A level maths
ok.. lets see if I got the "facts" straight... 40 degrees means that you turn your head 40 degrees and see the other airplane right? and the distance between you and that other airplane is decreasing at a certain rate ..... which means that the "hypotenuse" of our right triangle is getting shorter... does that seem about right?
yep thats how I see it, but not how I did it. I couldnt figure out how to do it properly so I just used the time tit travelled a mile while is 3 mins, assuming it is moving at a uniform speed but I could be wrong
ok.... then we know one simple fact. The planes appear to be approaching each other at a rate of 2 miles every 6 minutes... or 1 mile every 3 minutes. 10miles - 9 miles = 1 mile apart. How long does it take to fly 9 miles? 9*3 = 27 minutes to cover 9 miles... is that right?
err... 3 minutes to cover 9 miles lol
ack!!!... i confused myself again
27 minutes... 9 miles * 3 minutes per mile = 27 minutes.. lets stick with that
tell me when im wrong.... cause i can be wrong :)
lol yeah
thats what I got
then I added 27 to the original time but I dont know if thats right cos I dont have the answers T_T
10.50 + 0.27 =... 10.77 77 - 60 = 17 11.17 sounds good
Ok thanks. Btw how would you do it in calculus?
how would I do it in calculus? probably with alot of erasing :)
lol
youd have to know how fast each plane was traveling and the headings they were on to calculate the rates of change, so I dont think there is anough information to use calculus on this problem...
oh cool .. well thanks for your help. Gonna do some physics now. later
youre welcome :)
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