YumYum247
  • YumYum247
someone please help me?!?!?!?!?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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YumYum247
  • YumYum247
YumYum247
  • YumYum247
How do i figure out the distance and displacement???? :(
YumYum247
  • YumYum247
but before that please check if my drawing is ok???? :)

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YumYum247
  • YumYum247
@Michele_Laino
YumYum247
  • YumYum247
@dan815
dan815
  • dan815
wow u went all out with ur measuring lol
YumYum247
  • YumYum247
dan you are prally that missing brother i always wanted.... :")
YumYum247
  • YumYum247
when your marks are dangling by a single branch.....you have to XD
arindameducationusc
  • arindameducationusc
Displacement= Final position-initial position
arindameducationusc
  • arindameducationusc
for distance, just add 14+6+12+2
YumYum247
  • YumYum247
but dear there are directions that are opposite to each other and even more weird.... E, N, NE, NE
arindameducationusc
  • arindameducationusc
Distance doesnot depend on direction! Only magnitude
YumYum247
  • YumYum247
how do i subtract out the two vector direction from a one dimentional???
arindameducationusc
  • arindameducationusc
Ya, displacement is little complicated here.... give me some time....
arindameducationusc
  • arindameducationusc
For distance, just add (simple)
YumYum247
  • YumYum247
sorry i was referring to displacement......!!!
YumYum247
  • YumYum247
:D
arindameducationusc
  • arindameducationusc
okay....
YumYum247
  • YumYum247
so what's up with displacement???? ?XD
YumYum247
  • YumYum247
can someone please check my drawing if i've make it right so we can proceed with the question????
arindameducationusc
  • arindameducationusc
Trignometric functions cannot be used here... So, If you have taken accurate scale drawings, The method is absolutely correct.
YumYum247
  • YumYum247
that's right trig can't be used here since it's not a right triangle....
arindameducationusc
  • arindameducationusc
This is a pure scaling and drawing problem... Yup you are right..... Yes because no 90 degrees
arindameducationusc
  • arindameducationusc
Good job!
YumYum247
  • YumYum247
XD
YumYum247
  • YumYum247
the thing is that there are three triangles in this problem...XD
arindameducationusc
  • arindameducationusc
If protector is allowed in this question, taking angles and taking length of the distance gives answer.... which you have done....
YumYum247
  • YumYum247
displacement is my main concern here.....
arindameducationusc
  • arindameducationusc
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?
YumYum247
  • YumYum247
@joeyboy111
anonymous
  • anonymous
the question was answered why did you tag me?
YumYum247
  • YumYum247
@joeyboy111 how do i find the displacement????
anonymous
  • anonymous
take the final point and the initial point(in this case 0) and take the difference
YumYum247
  • YumYum247
what!!!! :O
johnweldon1993
  • johnweldon1993
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
johnweldon1993
  • johnweldon1993
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"
YumYum247
  • YumYum247
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) = ?
YumYum247
  • YumYum247
i meant right triagle****
arindameducationusc
  • arindameducationusc
@johnweldon1993 for 12km, the angle should be 75
arindameducationusc
  • arindameducationusc
and for 2km, angle should be 25
arindameducationusc
  • arindameducationusc
Yes trig can be used.. that was cool... @johnweldon1993
johnweldon1993
  • 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|
johnweldon1993
  • johnweldon1993
So for that reasoning, is why I'm using "sin" as the function for 'east' and "cos" for 'north' using those angles
arindameducationusc
  • arindameducationusc
@johnweldon1993 Can I give my argument here if you dont mind?
johnweldon1993
  • johnweldon1993
Yeah absolutely, that's why I said It's late it could be wrong. Always open to criticism
arindameducationusc
  • arindameducationusc
|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
johnweldon1993
  • johnweldon1993
Correct
arindameducationusc
  • arindameducationusc
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 ?
johnweldon1993
  • johnweldon1993
Do you agree that every vector has both a 'x' and 'y' "and z irrelevant here" component?
YumYum247
  • YumYum247
oh my god, so do you use the x component and y component way????? i know how do to that way....
arindameducationusc
  • arindameducationusc
See the red line here, I guess this should be the displacement, ofcourse they have component x and y..... give me one minute ........ hmmmm
1 Attachment
johnweldon1993
  • johnweldon1993
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 \[\large Displacement = Red Line = \text{How far over}^2 + \text{How far up}^2\]
arindameducationusc
  • arindameducationusc
yes.....
YumYum247
  • YumYum247
Thank you bunch of lumps of flesh and pile of dead skin cells, i really appreciated you help.... :")
johnweldon1993
  • johnweldon1993
haha no problem...let us know what you end up with!
YumYum247
  • YumYum247
sure will. :")
arindameducationusc
  • arindameducationusc
We had a blackout... Sorry couldnot reply k, bye....
YumYum247
  • YumYum247
YumYum247
  • YumYum247
YumYum247
  • YumYum247
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. :")
YumYum247
  • YumYum247

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