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May 6, 2015

May 6, 2015

Posted by **Chad** on Friday, April 12, 2013 at 12:43pm.

Hi, I am completely new to this, and missed a lot of classes due to hospital trips. Trying to catch up but there is no sample question in my text like this one. I dont really know how to get the numbers to make the vectors. Can you lead me in the right direction, would be very much appreciated, thank you very much.

The answer they get is 60.5 degrees north or 29.5 degrees north of east, but they show no work, just the answer. When I draw out the vectors, what do I use to determine the length of each vector? The hours? The knots? What scale 1mm = 1hour, or 1mm = 1knot?

Steve was nice enough to give me a hand, but me not being the smartest one on the block is still a little confused. Thank you Steve, very much appreciated!!! This site is the best.

- physics -
**Damon**, Friday, April 12, 2013 at 1:28pmMath books drive me crazy with their misuse of navigational language. You travel on a heading as used in the second sentence. A bearing is what direction you look to see something. (The enemy is on a bearing of 185 degrees true.)

Anyway start at (x,y) = (0,0)

It is heading 25 deg east of north

in x-y coordinates that is 25 degrees clockwise from the Y axis so we would say

X velocity = 4 sin 25 = 1.69 knots

Y velocity = 4 cos 25 = 3.625 knots

after three hours x = 3*1.69 = 5.07 nautical miles

and y = 3*3.625 = 10.875 nautical miles

Now we turn south (negative y axis)

go south for two hours at 5 knots = 10 nautical miles

so in the end after 5 hours

x position = 5.07 naut miles

y position = 10.875-10 = .875 naut miles

so that position on you x-y graph is

(5.07, .875 )

tan angle up from x axis = y/x =.875/5.07

so angle up from x = 9.79 degrees

so angle clockwise from north = 90-9.79 = 80.2 degrees east of north.

If you want it in km rather than nautical miles, multiply all distances by 1.852

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**Chad**, Friday, April 12, 2013 at 2:06pmSorry the answer they have is 60.5 degrees east of of north or 29.5 degrees north of east. There would be two vectors right, so do you take the degrees from the resultant? Thanks Damon, I do not nothing about this stuff at all. The question says my answer should be found by drawing, so it may not match.

- physics -
**Chad**, Friday, April 12, 2013 at 2:08pmI gave the wrong answer at first. Sorry.

- physics -
**Anonymous**, Friday, April 12, 2013 at 8:12pmThanks Damon, figured it all out thanks to your people's help, you rock. Thanks buddy!!!