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Vector subtraction question

  1. Sep 5, 2006 #1
    I'm in gr. 12 physics right now(just started school today), and we started some review work. This is a new school, so I'm guessing some of the things taught by the different teahers vary, which explains why I don't understand this question. Here's the question word for word, there's also a small amount of information before the question, which I'll also post:

    The vector subtraction A - Bis defined as the vector addition of A and -B, where -B has the same magnitude as B, but is in the opposite direction. Thus A- B = A + (-B).

    Ex 4. Given that A = 35m/s [27 degrees N of E] and B = 47 m/s [E], determine the change in velocity C = A - B.


    Now, I'm only a high school physics student, but the way I know to calculate a change in velocity is v_2 - V_1. This question threw me off completely, and I have no idea what I should do. The only thing I came up with, was this, and I'm sure it's wrong, but I'll post it anyway:

    C = A - B
    = 35 - 47
    = -12

    Therefore the change in velocity is -12 m/s.

    I don't think this can be right. Number one, the velocity is increasing from A to B, so I don't think the answer should be negative. And number two, this says nothing about the change in direction, only speed.

    Someone please help soon, thanks in advance.
     
  2. jcsd
  3. Sep 5, 2006 #2

    Chi Meson

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    You have not taken direction into account.

    A is 35 m/s, 27 degrees n of e.
    Find the componant north and the componant east.
     
  4. Sep 5, 2006 #3
    Ok, so I should find the North and East components of the vector, and then use that in the equation instead?
     
  5. Sep 5, 2006 #4

    Hootenanny

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    Yes, sounds good to me. Decompose both vectors into their North and East components find the change in each case, then recombine the two components to find the resultant vector. Don't forget to find and quote the bearing.
     
  6. Sep 5, 2006 #5
    What does that mean?

    Also, after I decompose the vector into it's components, and add them, what do I do? To find the new resultant vector, do I use the new East component, along with the previous North component to find the resultant vector?
     
    Last edited: Sep 5, 2006
  7. Sep 5, 2006 #6

    Hootenanny

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    Once you have performed the vector subtraction you will be left with a vector that has two components, North/South and East/West. Now, to fully answer the question, you must find the resultant vector. Now, you can find the magnitude using pythag; then you must use trig to find the angle or direction of this vector. Do you follow? I am assuming that you have done this before.
    Yes, this is totally acceptable because you will also be adding the north components together, you aren't leaving any out. For example, if I has to vectors and their components were (N= North and E = East);

    [tex]\vec{A} = 5N + 10E[/tex]

    [tex]\vec{B} = 4N - 6E[/tex]

    Then, the vector [itex]\vec{AB}[/itex] (or C = A + B) would be;

    [tex]\vec{AB} = (5+4)N + (10-6)E[/tex]

    [tex]\vec{AB} = 9N + 4E[/tex]

    Do you follow?
     
  8. Sep 5, 2006 #7
    I think I understand, now. I also editted my earlier post, because I started to understand what you were saying, a little better.

    So basically I add the North/South components, and the East/West components, and that gets me the resultant vector?
     
  9. Sep 5, 2006 #8

    Hootenanny

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    Yes, but then you need to combine the two components into a single vector, you can do this using pythag. Once, you have done this you need to find the angle and quote a bearing (e.g. 32o anti-clockwise from north or whatever it may be).
     
  10. Sep 5, 2006 #9
    Oh, alright. It seems a bit easier than I thought. It was just a familiar question masked in unfamiliar terms, it seems. I'll post my work when I'm done, so could someone make sure I did it right?
     
  11. Sep 5, 2006 #10

    Hootenanny

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    I find that I'm always a bit rusty after the breaks, but it soon comes flooding back :smile:
     
  12. Sep 5, 2006 #11
    Ok, I think I got the right answer, it's just the resultant vector of the two, right? But I was wondering something, what exactly is the resultant vector? What is the importance of this value? I know that in the case of displacement, the resultant vector shows the overall distance travelled, and in which direction, but this isn't displacement, this question is about velocity. What exactly does the resultant vector represent? The final velocity is still 47 m/s [E], isn't it?

    Also, I was wondering, what does this problem have to do with vector subtraction? From what I gathered, vector components were being added, but the explanation before the problem made things seem a little confusing,
     
  13. Sep 5, 2006 #12
    Someone please help.
     
  14. Sep 5, 2006 #13
    Vector subtraction is kind of a fancy way of saying that you can add two vector distances even if they are going opposite directions. The subtraction part just means that they want you to look at which direction the vectors are going. The resultant vector is the addition of the the two components. I think the resultant vector in this case is just used to help you find the average velocity. Hopefully this helps I am new to physics too so I may not 100 percent correct.
     
  15. Sep 5, 2006 #14
    Hmm, you make a good point, thanks for your help. I guess the resultant vector is just the average velocity, I think that works, at least in a simple case. Can anyone else clarify?
     
  16. Sep 6, 2006 #15

    andrevdh

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    Initially the object moved in the direction of vector A. Now it is moving in the direction of vector B. The change in it's velocity is given by vector C, that is vector C is the difference between the two velocity vectors or it is the sum of vector A and -B. Vector C therefore tells us by how much and in which direction the object's velocity had to be changed in order to move with the new speed in the direction of vector B.
     

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    Last edited: Sep 6, 2006
  17. Sep 6, 2006 #16
    I'm guessing that the attachment in andrevdh's post is a diagram of some sort. I appreciate the help, but, while I'm waiting, is there anything else anyone could say or show to help clarify this? andrevdh said that C represented the change between A and B, but I still don't understand some things. Why does the formula use A and -B? Why do you make B negative? And how exactly does the resultant vector represent the change in velocity? Sorry if this is really simple, but it's a new school, and some of these terms/concepts I never learned before, so I'm a bit confused.
     
  18. Sep 6, 2006 #17

    andrevdh

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    With ordinary quantities it works the same way:

    If you had one hundred dollars (A) and you now have 40 (B) then the original amount was changed by 100 - 40 = 60 (C). That is the difference between the "old amount" and the "new amount" gives the change in the quantity. This relationship that the original amount is conserved is not necessary applicable in the case of vectors. A force applied to the object can alter it's velocity in both magnitude and direction, resulting in a larger or smaller velocity than originally.
     
  19. Sep 6, 2006 #18
    Ok, but I have another question, sorry if it was already answered indirectly or something, but it's kind of confusing to me. Say an object travels 5 m/s [N], and then 5 m/s , would you say it has no velocity, in this case? I just find it confusing, because in reality, the object is still moveing 5 m.s , but for some reason, I don't think this is the case here.
     
  20. Sep 6, 2006 #19

    Hootenanny

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    No, intially it would have a velocity of 5 m.s-1 N. Then after an acceleration it would have a final velocity of 5 m.s-1 S. At no point is the velocity zero.
     
  21. Sep 6, 2006 #20
    Why were there exponents with your velocities?

    And what about when you add the vectors? You get a result of 0 m/s. What does this zero represent?
     
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