1. YumYum247

2. YumYum247

How do i figure out the distance and displacement???? :(

3. YumYum247

but before that please check if my drawing is ok???? :)

4. YumYum247

@Michele_Laino

5. YumYum247

@dan815

6. dan815

wow u went all out with ur measuring lol

7. YumYum247

dan you are prally that missing brother i always wanted.... :")

8. YumYum247

when your marks are dangling by a single branch.....you have to XD

9. arindameducationusc

Displacement= Final position-initial position

10. arindameducationusc

11. YumYum247

but dear there are directions that are opposite to each other and even more weird.... E, N, NE, NE

12. arindameducationusc

Distance doesnot depend on direction! Only magnitude

13. YumYum247

how do i subtract out the two vector direction from a one dimentional???

14. arindameducationusc

Ya, displacement is little complicated here.... give me some time....

15. arindameducationusc

16. YumYum247

sorry i was referring to displacement......!!!

17. YumYum247

:D

18. arindameducationusc

okay....

19. YumYum247

so what's up with displacement???? ?XD

20. YumYum247

can someone please check my drawing if i've make it right so we can proceed with the question????

21. arindameducationusc

Trignometric functions cannot be used here... So, If you have taken accurate scale drawings, The method is absolutely correct.

22. YumYum247

that's right trig can't be used here since it's not a right triangle....

23. arindameducationusc

This is a pure scaling and drawing problem... Yup you are right..... Yes because no 90 degrees

24. arindameducationusc

Good job!

25. YumYum247

XD

26. YumYum247

the thing is that there are three triangles in this problem...XD

27. arindameducationusc

If protector is allowed in this question, taking angles and taking length of the distance gives answer.... which you have done....

28. YumYum247

displacement is my main concern here.....

29. 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?

30. YumYum247

@joeyboy111

31. anonymous

the question was answered why did you tag me?

32. YumYum247

@joeyboy111 how do i find the displacement????

33. anonymous

take the final point and the initial point(in this case 0) and take the difference

34. YumYum247

what!!!! :O

35. 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

36. 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"

37. 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) = ?

38. YumYum247

i meant right triagle****

39. arindameducationusc

@johnweldon1993 for 12km, the angle should be 75

40. arindameducationusc

and for 2km, angle should be 25

41. arindameducationusc

Yes trig can be used.. that was cool... @johnweldon1993

42. 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|

43. johnweldon1993

So for that reasoning, is why I'm using "sin" as the function for 'east' and "cos" for 'north' using those angles

44. arindameducationusc

@johnweldon1993 Can I give my argument here if you dont mind?

45. johnweldon1993

Yeah absolutely, that's why I said It's late it could be wrong. Always open to criticism

46. 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

47. johnweldon1993

Correct

48. 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 ?

49. johnweldon1993

Do you agree that every vector has both a 'x' and 'y' "and z irrelevant here" component?

50. YumYum247

oh my god, so do you use the x component and y component way????? i know how do to that way....

51. 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

52. 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 $\large Displacement = Red Line = \text{How far over}^2 + \text{How far up}^2$

53. arindameducationusc

yes.....

54. YumYum247

Thank you bunch of lumps of flesh and pile of dead skin cells, i really appreciated you help.... :")

55. johnweldon1993

haha no problem...let us know what you end up with!

56. YumYum247

sure will. :")

57. arindameducationusc

58. YumYum247

59. YumYum247

60. 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. :")

61. YumYum247