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Average velocity vector question. Try again

  1. Feb 16, 2007 #1
    1. The problem statement, all variables and given/known data
    A train with constant velocity of 60km/hour travels east for 40minutes; then in a direction 50degrees east of north for 20minutes; then west for 50minutes. A)What are the magnitude and B) angle of its average velocity during the trip?



    2. Relevant equations sinx; cosx; tanx<-- I think.



    3. The attempt at a solution
    My main trouble is finding out what the question is asking. I have determined that the magnitudes of each are 40km;20km;50km respectively.
    I have drawn each vector on a cartesian coordinate plane such that they start at the origin and then connect tip to tail. IS the displacement vector the one that extends from the origin to the tip of the last one drawn?

    This is my first post so I apologize if it is difficult to follow. Is there a way I could have drawn this w/out a scanner? Thanks.
     
  2. jcsd
  3. Feb 16, 2007 #2

    Dick

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    Science Advisor
    Homework Helper

    Your description is clear enough. And yes, the total displacement is just the vector connecting the starting point with the ending point.
     
  4. Feb 16, 2007 #3
    Here is the work I have done, but the book is giving a solution of 7.9km/h....and 22.5 degrees east of north. I dont know where I went wrong. Am I approaching this all wrong?


    [​IMG]
     

    Attached Files:

    Last edited: Feb 16, 2007
  5. Feb 16, 2007 #4

    berkeman

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    Staff: Mentor

    The magnitude of the average vector velocity is the magnitude of the final displacement vector divided by the total time (and convert units to m/s as needed). The angle is the angle of your resultant V up from the horizontal (East) axis. I get about the right answer using the numbers from your picture -- did you account for the time correctly?
     
  6. Feb 16, 2007 #5
    sorry...the solution was not in m/s...it is 7.9km/h.....The displacement vector is the dotted life correct? which equals 19.9. Divided by by 1 and 5/6 of an hour is getting me 10.8.
    And the angle is stated as East OF North, meaning the angle formed by the Y axis and the vector(clockwise).
     
  7. Feb 16, 2007 #6

    berkeman

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    Staff: Mentor

    I think I see your error. See how it's initially 40+15.3km to the right, and then later 50km to the left? I had wondered where you were getting the 19 number from.....
     
  8. Feb 16, 2007 #7
    19 is the length of hypoteneuse (the displacement vector) formed by the right triangle with a=15.3 and b=12.8.
     
  9. Feb 16, 2007 #8
    Wow. You are correct in that regard. 55.3 east NOT 65.3.....Holy oversight Batman. ...........I am still getting errors, but it is extremely close now; I am sure they are round-off errors. I will re-calculate the problem. Thank you.
    ~Casey
     
  10. Feb 16, 2007 #9
    Much Better.:cool:

    [​IMG]

    These evaluations yield the correct solutions.:smile:
     
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