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AP Physics: Vectos and their properties

  1. Oct 1, 2008 #1
    5. A plane flies from base camp to Lake A, a distance of 280 km at a direction of 20.0 degrees north of east. After dropping off supplies, the plane flies to lake B, which is 190 km and 30.0 degrees west of north from Lake A.

    I'm not even going to write the actual question part, since that's not the issue I'm having. Basicaly, I'm having trouble drawing the vector 30.0 degrees west of north. :bugeye: I know how to draw a vector North or south of East or west, but the reverse of that can't get through my head. I tried, and I ended up getting 225 km from Lake B to the camp, but the key says 310. My scale is set at 1 cm=50 km, and my vector for camp to Lake A is 5.5 cm (about) at the correct angle. Maybe just a picture to tell me the correct direction?

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    Just to check a second question:

    2. An airplane flies 200 km due west from city A to city B and then 300 km in the direction of 30.0 degrees north of west from city B to city C. In striaght-line distance, how far is city C from city A and in what direction is City C in relation to city A?

    485 km 20 degrees N of W.
     
  2. jcsd
  3. Oct 1, 2008 #2
    "Basicaly, I'm having trouble drawing the vector 30.0 degrees west of north."
    Draw a vertical vector... then draw another vector 30degrees to the left of the first one.

    For the 2nd one.. it seems about right. Don't really have time right not to put in calculations and better explanations.. hopefully this will help.
     
  4. Oct 1, 2008 #3
    The first position vector is 280 long, 20 degrees North of East. If you can do that one then, its not so hard to draw from Lake A, the second vector, 190 long, in a direction 30 degrees West of North.

    Draw a thin reference line North-to-South going right through the point at Lake A.
    Imagine you are at Lake A, facing North. Then 30 degrees West of North means turn 30 degrees towards the West, ie. anti-clockwise.

    OK - You can do it graphically, with everything to scale.

    You COULD instead, use sines and cosines to figure the total distance moved Northwards, and also the final distance moved Eastwards, and then use Pythagoras theorem to get the distance from Camp to Lake B.
     
  5. Oct 1, 2008 #4
    We're only supposed to use the graphical method for this, but I think this really helps. Thanks!
     
  6. Oct 2, 2008 #5
    Thats OK - the teacher need not know you were able to check it out another way.

    When you have drawn it, you can see a triangle with the hypotenuse 280 long, and 20 degrees at the base camp. The North-South line meets East-West line line 90 degrees. That makes the remaining angle 70 degrees.

    Then, after the 190 line to Lake B is drawn, you can see a new triangle with the 30 degrees in it.
    That leaves 180 - 70 - 30 = 80 degrees angle between the arriving 280 vector, and the departing 190 vector.
    If you complete that last triangle by drawing the final line between base camp and Lake B, you can measure it and get something near 310.

    Notice that you have a triangle with two sides known, and also the angle between them.
    The well known formula for a triangle with sides a, b, c, and angles opposite A, B, C is a2 = b2 + c2 - 2*b*c*cos(A)
    so get a calculator and do 2802 + 1902 - 2*280*190*cos(80)
    Then find the square root of the above. Thats the accurate answer. Its faster than fiddling with scales!
     
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