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Homework Help: Vector Direction (west of North)

  1. Sep 20, 2006 #1
    Hi, my first post.. i am taking an intro physics class at uni and its been awhile since I've taken highschool physics.

    My question is if a car is moving 25 degrees west of North what direction is that? Is that the same thing as moving 155 degrees north of west?

    How do I draw the vector?

    Thanks for your help!
  2. jcsd
  3. Sep 20, 2006 #2
    it is the same as 65 degrees North of West
  4. Sep 20, 2006 #3
  5. Sep 20, 2006 #4
    okay so here is the question:
    a car is driven 25 degrees west of north with a speed of 6.5km/h for 15 min. then due east with a speed of 12km/h for 7.5 min the car completes the final leg in 22 min. What is the final direction and speed of her travel on the final leg (assuming her speed is constant and the car returns to its starting point at the end).
    so I've drawn my vectors and I know it is heading west of south but I don't know how to approach this question.
    The text I am using is James Walker 3 edition and the examples are absoluetely of no help!

    I tried doing it and I got 2.7km/h ( but I don't think it is right)
    How should look at this?
  6. Sep 20, 2006 #5
    I did it again and this time I got 10.97km/h 32.5 degrees west of south.(or south of west?)..does that look right?
    Last edited: Sep 20, 2006
  7. Sep 20, 2006 #6

    Chi Meson

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    Take out the "of" and substitute the word "from."

    "West of North" becomes "west from north." That is start with "due north" and rotate the required number of degrees to the west. You have gone west, from north.

    I got 61 degrees south of west, (close to your answer) but the speed I calculted was less than half of what you got.
  8. Sep 20, 2006 #7
    okay I still don't get it.....here is what I did:

    I treated it as two seperate triangles: (can I do that?)
    12-2.747=9.25 km/h
    6.5cos(25) =5.89km/h
    then c2=a2+b2
  9. Sep 20, 2006 #8
    i calculated 4.58 km/h and 61.1 degrees south of west.

    i used the law of cosines

    c^2 = a^2 + b^2 - 2abcos C
    c^2 = (1.5^2) + (1.625^2) - (2)(1.5)(1.625)cos 65
    c = 1.68 km

    speed = 1.68*(60/22)
    speed = 4.58 km/h

    to get the angle i did...

    1.625*sin(25) = .687 km (distance from end of first vector to y axis)
    1.5-.687 = .813 km ( horizontal distance from y axis to end of 2nd vector)

    sin(x) = .813/(1.68)
    x = 28.9 degrees west of south or 61.1 degrees south of west
  10. Sep 20, 2006 #9

    instead of using velocities in your calculations try using distances by taking the velocity and multiplying it by the amount of time the person traveled. For instnace: 6.5 km/h * (15/60) = 1.625 km
  11. Sep 20, 2006 #10
    oops sorry didn't read the above post thanks so much that makes more sense
    Where did you get those values? (1.5 and the 1.635 from?)
    60/22 what are those values?

    to get the angle i did...

    Sorry physics seems to be an extreme weakness for me!
    Thanks so much for your help :bugeye:
  12. Sep 20, 2006 #11
    1.5 is when you take 12km/h and multiply it by (7.5/60) . since the person traveled 7.5 mins you convert that into hours (7.5/60) which is .125 hours. you multiply that by 12km/h to get the distance traveled (1.5km). use this method to use distances instead of velocities in your calculations

    60/22 is simply doing the converion in reverse since i got a distance value and i wanted to turn it back into a velocity (remember 22 is the number of minutes the person traveled on the last leg).

    your methods seem correct you just need to use distances instead of velocities
    Last edited: Sep 20, 2006
  13. Sep 20, 2006 #12

    I get it now.. I tried both ways once treating it as two seperate triangles and the second using the cosine law and both worked ... and the angle was found too..

    thanks again
  14. Sep 21, 2006 #13
    Remember, you can solve this GRAPHICALLY! As long as you draw everything to scale and use a protractor, you can check your answer.
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