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Average Velocity. What Am I doing wrong

  1. Sep 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A train at a constant 42.0 km/h moves east for 33 min, then in a direction 64.0° east of due north for 25.0 min, and then west for 69.0 min. What are the (a) magnitude (in km/h) and (b) angle (relative to north, with east of north positive and west of north negative) of its average velocity during this trip?

    2. Relevant equations
    The way I inturpreted the problem I needed to first find the time in hours and the displacement in kilometers for each vector then I added the displacements over the times. here is the work I did.


    3. The attempt at a solution
    First I calculated the total time. Since time isn't a vector I added everything.
    33'*/60'=.550h
    25'*/60'=.417h
    69'/60'=1.15h
    Sum = 2.12h

    Next I found the total distance
    42km/h*.550h=23.1km
    42km/h*.417h=17.5km
    42km/h*1.15h=48.3km

    then I added the magnitudes of the three vectors taking the magnitude of the x coordinate for the second
    r=(23.1km)+(17.5km *Cos(26))-(48.3km)= -9.47

    Finally I divided displacement over time to get an unconvincing and completely wrong answer of -4.47

    What did I do wrong.
     
  2. jcsd
  3. Sep 26, 2009 #2
    For average velocity, it should be sufficient to consider the total displacement (resultant vector of the sum of all displacement vectors). Correct me if I'm wrong.
     
  4. Sep 26, 2009 #3
    You need to sort out the difference between average speed and average velocity.
     
  5. Sep 27, 2009 #4

    Redbelly98

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    Homework Helper

    .
    Welcome to Physics Forums.
    Looks good so far.

    This is just the x-coordinate of r. You need to also:

    1. Use the angle w.r.t. the x-axis when you take the cosine to get the x-component.
    2. Find the y-coordinate of r, so that you get the complete displacement vector r.
     
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