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2-D Vector Problem

  1. Apr 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Darryl drives his load of tomatoes 14.0 km [E], 6.0 km [N], 12.0 km [ N 15° E], and then 2.0 km [N 65° E]. This takes him 42 minutes. Calculate Darryl's distance and displacement.
    a) Calculate Darryl's distance and displacement. Draw a diagram and show your work.
    b) Calculate Darryl's average speed and average velocity (record your answer in m/s).

    2. Relevant equations
    C^2 = a^2 + b^2
    c(squared)=a(squared) +b(squared)=2abcos∅

    3. The attempt at a solution

    so i've tried drawing this and it seems that i have 2 triangles. so i tried to solve for Δd on the first triangle using the cosine law. to get the angle i used Tan
    c(squared)=a(squared) +b(squared)=2abcos∅
    = 6(squared)+14(squared)-2(6)(14)cos23°
    [second triangle]
    tan∅=2km/12km = 9.4°
    c(squared)=a(squared) +b(squared)=2abcos∅

    so then i figured add them together and my Δd = 109.2 km

    now honestly this just doesn't seem right to me at all I'm not sure how i should be approaching this but i'm getting really flustered cause i'm completely stumped by vectors right now and it's starting to turn me away from physics.
    Last edited: Apr 3, 2012
  2. jcsd
  3. Apr 3, 2012 #2
    Since motion is in two directions, I would use the two vectors Dx and Dy. I would also set up the coordinate system such that the starting point was at the origin and that north is the +y direction, and east is the +x direction.

    Now to address the problem, in part a) I believe distance to be the amount he traveled (although honestly I may be wrong here, I can't seem to remember these terms right now). So since this only takes into account the scalar part of his movement, you can add all of the distances he traveled and get the distance he traveled, which in this case is 14 + 6 + 12 +2 = 34km.

    The displacement is a bit tougher, as this is comparing where he ended to where he started. This is where the vector math comes in handy.
    Starting with the first segment of his journey, 14.0 km [E], you can note that this motion is directly east, and only effects his displacement along the x-axis. Therefore you add 14 to Dx and make its value 14. Dy is still 0 because he hasn't moved north or south yet.
    Next, he travels 6.0 km [N], which leads me to add 6 to the variable Dy. Now Dx is 14 and Dy is 6.
    To handle the 12.0 km [ N 15° E] step, you need to break it up into north and east components. If you draw a triangle with its right angle in the top left, its two legs parallel with the x and y axis', and the hypotenuse being 12, you can note that the distance along the horizontal leg to be 12sin(15), and the distance along the vertical leg to be 12cos(15). Therefore you can say that Dx = 14 + 12sin(15) and Dy = 6 + 12cos(15)

    From here you do a similar process with the last segment of his travel, and then once you have a final value of Dx and Dy you can make one large triangle, with Dx being the bottom leg and Dy being the right leg, and from this triangle you can solve for the hypotenuse (displacement) and the angle of this displacement.

    Does this help at all? I can attempt to explain it differently of upload a few pictures if what I'm saying isn't making sense. Hopefully this addresses some of the issues you are having with vectors at least.
  4. Apr 4, 2012 #3
    okay wow did that ever help !!
    i think i finally figured it out.
    after doing what you said i had Dy = 18.4 and Dx = 10.91
    then i used c^2 = a^2+b^2 - 2abCocC
    = 11^2+18^2-2(11)(18)cos90degrees
    = 21.09
    then to find the angle i used sine
    sinB/b = sinC/c
    angle = 54.9 degrees
    so then darryls displacement would be 21km, 55 degrees east of north.
    then converted into metres of course which would be 171.4m 55 degrees east of north.

    or should i have just done c(squared)=a(squared)+b(squared) which would have been 11^2 + 18^2 = c^2 which still equals 21.09km

    and maybe for the angle i should have just used tan∅= 18/11 = 58°

    i may be wrong but this is what I've come up with hopefull I'm on the right track now
    Last edited: Apr 4, 2012
  5. Apr 4, 2012 #4
    for the average speed and velocity i did

    Vav→ = 171.4m[N55°E] / 42mins (2520sec) = 0.068m/s [N55°E] or 4.08m/m
    metres per minute make more sense to me but it wants metres per second right but for the next part seconds just didn't seem to produce a correct answer

    Vav = 4.08m[N55°E]/ 42mins (2520sec) = 0.097 m/m or 0.0016m/s

    im a tad bit confused as those don't seem correct
  6. Apr 4, 2012 #5


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    Staff Emeritus
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    Avg speed = distance/time

    (6000 m + 12,000 m + 2000 m)/(42 min * (60 s/min))

    Make sense?
  7. Apr 4, 2012 #6
    okay well my total distance was 34km and total time was 42mins
    so that was 0.809 km/min which would be 13.48m/s
  8. Apr 5, 2012 #7


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    Staff Emeritus
    Science Advisor
    Gold Member

    Yeah that sounds good. I totally didn't see that first distance of 14 km when I read the problem before! Hence why it was missing from my example above. :wink:

    For average velocity, you need to use displacement rather than distance.
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