someone please help me?!?!?!?!?

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someone please help me?!?!?!?!?

Physics
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How do i figure out the distance and displacement???? :(
but before that please check if my drawing is ok???? :)

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wow u went all out with ur measuring lol
dan you are prally that missing brother i always wanted.... :")
when your marks are dangling by a single branch.....you have to XD
Displacement= Final position-initial position
for distance, just add 14+6+12+2
but dear there are directions that are opposite to each other and even more weird.... E, N, NE, NE
Distance doesnot depend on direction! Only magnitude
how do i subtract out the two vector direction from a one dimentional???
Ya, displacement is little complicated here.... give me some time....
For distance, just add (simple)
sorry i was referring to displacement......!!!
:D
okay....
so what's up with displacement???? ?XD
can someone please check my drawing if i've make it right so we can proceed with the question????
Trignometric functions cannot be used here... So, If you have taken accurate scale drawings, The method is absolutely correct.
that's right trig can't be used here since it's not a right triangle....
This is a pure scaling and drawing problem... Yup you are right..... Yes because no 90 degrees
Good job!
XD
the thing is that there are three triangles in this problem...XD
If protector is allowed in this question, taking angles and taking length of the distance gives answer.... which you have done....
displacement is my main concern here.....
Ofcourse. See the idea is after taking all those angles and carefully going the path, take the final point and the initial point(in this case 0) and take the difference. Got it?
the question was answered why did you tag me?
@joeyboy111 how do i find the displacement????
take the final point and the initial point(in this case 0) and take the difference
what!!!! :O
Displacement, Hmm, I know above we said trig cannot be used...but I cannot see why not? The trip looks like |dw:1438831247053:dw| If we think about "How far total East did he travel?" and "How far total North did he travel" Then we just have 1 giant right triangle with 2 sides known East...14 + 12sin(15) + 2sin(65) = ? North...6 + 12cos(15) + 2cos(65) = ? It's late so let me know if this is totally wrong
Also yes I know my 15 degree angle and 65 degree angled vectors should be swapped...but the equations remain the same 15 should be much more "north" and the 65 should be much more "east"
now worries about the angles, just tell me once i have the followings figured out, then i make a triangle and solve for the hypothenuse ? East...14 + 12sin(15) + 2sin(65) = ? North...6 + 12cos(15) + 2cos(65) = ?
i meant right triagle****
@johnweldon1993 for 12km, the angle should be 75
and for 2km, angle should be 25
Yes trig can be used.. that was cool... @johnweldon1993
Well the angles I'm using North 15 degrees east...I'm interpreting as it would be north...but instead its just 15 degrees east of north Thus the reason I'm labeling |dw:1438832198665:dw|
So for that reasoning, is why I'm using "sin" as the function for 'east' and "cos" for 'north' using those angles
@johnweldon1993 Can I give my argument here if you dont mind?
Yeah absolutely, that's why I said It's late it could be wrong. Always open to criticism
|dw:1438833553890:dw| See the black line I have drawn, As per the definition of displacement, we need that measurement. So, we need the length of the hypotenuse, i.e14^2+6^2
Correct
So, as per the displacement definition, if you see @YumYum247 's diagram. The triangles are not in 90 degrees. only the first hypotenuse can be used, not others as per the definition. What do you say @johnweldon1993 ?
Do you agree that every vector has both a 'x' and 'y' "and z irrelevant here" component?
oh my god, so do you use the x component and y component way????? i know how do to that way....
See the red line here, I guess this should be the displacement, ofcourse they have component x and y..... give me one minute ........ hmmmm
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That should INDEED be the displacement My approach is to find how far over "x" and how far up "y" in order to find that red line being the hypotenuse \[\large Displacement = Red Line = \text{How far over}^2 + \text{How far up}^2\]
yes.....
Thank you bunch of lumps of flesh and pile of dead skin cells, i really appreciated you help.... :")
haha no problem...let us know what you end up with!
sure will. :")
We had a blackout... Sorry couldnot reply k, bye....
can someone please check my work, i had to figure out the average velocity and average speed too, see if i've done it correct. :")

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