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Vectors and Distance

  1. Sep 9, 2004 #1
    An explorer is caught in a whiteout while returning to camp. He was supposed to travel due north for 5.0 km, but when the snow clears, he discovers that he actually traveled 7.8 km at 48° north of due east.

    (a) How far must he now travel to reach base camp?

    (b) In what direction must he travel (counterclockwise from due east)?
    I thought I understood what to do, but apparently I don't since I keep getting the wrong answer. how do I start this problem and how do I figure out part b.
  2. jcsd
  3. Sep 9, 2004 #2


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    You didn't say what you actually tried but here's a hint:

    [tex]x = d \cos \theta[/tex]
    [tex]y = d \sin \theta[/tex]
  4. Sep 9, 2004 #3
    I know to use the equations but I don't know what to plug in for the distance.
    I tried x=7.8sin48 and y=7.8cos48. I don't know if this is right and even if it is I don't know how to incorporate the 5.0 km and find out the distance he needs to travel to reach base camp (part a).
  5. Sep 9, 2004 #4


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    You need to find the x and y for the actual location AND the x and y for the desired location. I'm sure you have been given the formula for the distance between two points which you can use to obtain your final result.
  6. Sep 9, 2004 #5


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    This guy is originally at coordinates (0,0) and he wants to go to (0,5), but ends up at (x,y) instead. First calculate x and y, using the formulae that Tide suggested (with the angle 48 degrees). Then calculate (0,5)-(x,y). This is the vector that describes the last trip he has to make.
  7. Sep 9, 2004 #6
    I'm so lost. After working with the two equations I have x-5.2 and y=5.8. Now what do I do from here?
  8. Sep 9, 2004 #7


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    First of all, you should always keep at least one more decimal in intermediate results. The final answer should be given with 2 significant figures, but it's a bad idea to do the same with results that you intend to plug into another equation.

    So you've found that x=5.2192 and y=5.7965. That's a good start. As I said before, the next trip the explorer has to make is described by the vector (a,b)=(0,5)-(x,y). The magnitude |(a,b)| of that vector is the distance he has to walk. The direction of (a,b) is the direction he has to walk. When you've found a, b and |(a,b)|, you should be able to use the same equations as before to find the angle that identifies the direction.
  9. Sep 9, 2004 #8
    so my equation will look like
    (a,b)=(0,5)-(5.2192 ,5.7965)?

    how do i figure out a,b, i know this must be simple but I just want to make sure.
  10. Sep 9, 2004 #9
    Vector Analysis


    First, I would inform you that, right now, your calculations are simple, and while there are several places on the Internet that offer help. What you are presently dealing, because it is quite simple, my suggestion is that you try to construct a diagram of the problem; Graph the Problem / Draw a picture of what is going on. Then the resolution would be much easier.

    GingerBread27, there is so much that I could tell you about mathematics, but because most teachers of mathematics do not actually know or understand mathematics. If I were to begin this lesson, I would cause mose harm than good.

    Anyway, I hope this little bit helps.

    e. terrell
  11. Sep 9, 2004 #10
    This is just subtraction of two vectors.

    Originally you wanted to go 5km due north, but instead went 7.8 km, 48 degrees north of east. Well the answer we want is just the difference of these two vectors, the displacement( ie distance and direction).

    Now the 48 degrees is just 48 degrees above the x axis, if we take the east-west direction as our x axis, and likewise north as our positive y direction.

    Breaking up the 7.8km into components, we have 7.8cos(48) in x direction and 7.8sin(48) in y direction.

    We have to subtract the 7.8sin(48) from the 5 km, to find the total distance need to be travelled in the y to get to the destnation.

    5km - 7.8sin(48)km=5-5.7965 =-0.7965km. The negative means he should travel down or south to get to destination

    In x all we have is 0 -7.8cos(48)=-4.6839km. The minus means he should travel west

    Using pythogorean theorem, d^2=x^2+y^2, the total (distance)^2 is
    (-4.6839)^2+ (-.7965)^2= 22.5734=d^2

    d=4.7511 km

    the angle = arctangent(y/x)=arctan(-.7965/-4.6839)= 9.6509 degrees. However we are in the 4th quadrant because both the x and y components are negative, so we must add 180. ans = 189.65 deg.

    ans to
    (a) 4.75 km - don't hold me to it

    (b) 189.65 degrees counterclockwise from the east. don't hold me to it

    I believe this is correct. I hope this helps. :rolleyes:
  12. Sep 9, 2004 #11
    The answer to part A should be 5.3, I used the law of cosines to figure it out. Thank You all for your help. I am now working on part B.
  13. Sep 10, 2004 #12
    vectors and displacement

    Yes, I made an error.

    I used -4.6839. Sorry.

    (-5.219)^2+(-.7965)^2= 27.879312 =d^2

    d= 5.28

    The angle will be arctan(y/x)=arctan(-.7965/-5.219)= 8.677 deg
    But since we are in the 4th quad add 180 to ans, we get 188.7 deg from due east.

    or using law of sines: a/sinA=b/sinB

    I wish I could draw a picture, then things would be clear how you use law of sines.

    5.3/sin(90-48) =5/sinA => sinA =5sin(42)/5.3 = 0.6309947

    A=arcsin(.631)=39.1 deg, this is the angle between 5.3km and 7.8km.
    The angle between 7.8km and due south is 42 deg. 90 -(42+39.1) is the angle below west =8.9 deg

    Thus 180 + 8.9 is the angle from due east he must travel.=188.9 deg

    Just be careful to round correctly and to use one extra digit in your calculations to get the desired degree of precision. My numbers may be a little off because I used 5.3km instead of 5.219km. But the methods are correct. Understanding the method is what is important. I plucked numbers in so that you would believe me. If there is any question about this post or my previous post don't hesitate to ask.
    Once again sorry about the numbers in the previous post I made an error in punching in 7.8cos(48), the method is correct.

    While using the law of cosines and sines is perfectly legit and if anyone asked you this problem in every day life, using law of sine and cosine is in my opinion the easiest method.

    And may be the only method in your arsenal of tricks if you are in high school, but if this is a problem from a college text, then I think they may want you to start thinking in terms of components.

    Understanding how to add vectors by breaking them up into components is critical because this will be the easiest and fastest when you have more than three vectors, which in the so called real world problems, usually is the case. Dont ask what I mean by real world problems :smile:

    Trust me, get this component stuff under your belt, and you will be acing college physics I, and mechanical engineering problems with more ease than you would without mastering component stuff. :rofl:

    Once again method is correct, just double check my numbers.
    Last edited: Sep 10, 2004
  14. Sep 11, 2004 #13
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