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Trying to Understand Accel Vectors

  1. Sep 11, 2010 #1
    This is a sample problem in the text that I am having problems comprehending.

    1. The problem statement, all variables and given/known data

    A car is traveling northwest at 9.00 m/s. Eight seconds later it has rounded a corner and is headed north at 15.00 m/s. What are its magnitude and direction of its average acceleration during those eight seconds?

    2. Relevant equations

    ∆v = V_f - V_i (vectors)

    3. The attempt at a solution

    The problem shows its strategy, but wants the student to do the calculations for each step.

    Step 1: Write out V_i. Vector V_i=9m/s = -6.36m/s xhat + 6.36m/s yhat. I'm OK here. I know how to get velocity vectors and I worked this out correctly.

    Step 2: Write out V_f. Vector V_f = 15m/s. I guess this is just a given. OK ....

    Step 3: Calculate ∆V. ∆V= 6.36m/s xhat + 8.64m/s yhat. I don't know why the xhat is positive all of a sudden, but I understand that a turn due north adds zero to the xhat. I have no idea where the number for yhat is coming from.

    There's a step 4 and 5, but if I can understand this part, I think I can work those out. Thanks in advance.

    On a side note, how the heck can I type the symbols for hat, vectors, subscripts, etc.?
     
  2. jcsd
  3. Sep 11, 2010 #2
    I think the reason that the sign of the x component changes to positive is due to the fact that when you subtract the two vectors v_f and v_i, the signs of v_i changes so the x component becomes positive (becuase it was previously negative) and the y component becomes negative. So, since v_f has no x component in the first place, the resulting vector has the positive x component of the vector -v_i. The number for the y component appears to simply be 15 - 6.36 = 8.64, which is the result of subtracting the y components of v_f and v_i.

    With regards to typing mathematical symbols, see the thread here: https://www.physicsforums.com/showthread.php?t=8997
     
  4. Sep 11, 2010 #3

    PhanthomJay

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    Subtracting vector A from vector B is the same as adding vector B to the negative of vector A, tht is B - A = B + (-A). The negative of a a vector is a vector pointing in the opposite direction.

    (Edit: which is essentially what stevebac said).
     
  5. Sep 11, 2010 #4
    Working on it now .....
     
  6. Sep 11, 2010 #5
    OK, I'm still hopelessly lost, but here's what I've done to try and find the vectors:

    Vector A = Ax+Ay = 9m/s
    Ax = -6.36m/s
    Ay = 6.36m/s

    Vector B = Bx+By
    Bx = 0
    By = 15

    Vector C = Cx+Cy
    Cx = Ax+Bx = -6.36 + 0 = -6.36
    Cy = Ay+By = 6.36 + 15 = 21.36

    Vector C = (-6.36)² + (21.36)²; basic Pythagorean Theorem
    Vector C = 22.29

    So to calculate average acceleration over 8 seconds, I would think that I need to find ∆V/∆t, which would be 22.29/8, which equals 2.79 seconds, which is woefully incorrect. I could also see 6.36/8 sec to find the accel in the x direction and then figure out the accel in the y direction, but my 21.36 number is way off. I'm not understanding the neg/pos thing after reading what y'all wrote.

    I'm close though, and still battling ......
     
  7. Sep 11, 2010 #6
    Got it! Thanks for the help!!
     
  8. Sep 11, 2010 #7
    This part is incorrect. Vector C is the change in velocity, which means
    C = B - A

    Therefore,
    Vector C = Cx(i)+Cy(j)
    Cx = Bx - Ax= 0 - (-6.36) = 6.36
    Cy = By - Ax = 15 - 6.36 = 8.64

    Hence the magnitude of the velocity vector, which is the speed, is about 10.7

    Edit: glad to see that you've solved it.
     
    Last edited: Sep 11, 2010
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