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Calculate frank's distance and displacement.

  1. Jun 29, 2013 #1

    Zondrina

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

    Frank drives a load of tomatoes 14 km [E], 6 km [N], 12 km [N 15° E] ( East 15 degrees of north ) and then 2 km [N 65° E]. The trip takes 42 minutes.

    a) Calculate frank's distance and displacement.
    b) Calculate frank's average speed and average velocity.

    2. Relevant equations

    ##Δt = 42m##

    ##v_{av} = \frac{Δd}{Δt}## - ave speed

    ##\vec{v_{av}}= \frac{\vec{Δd}}{Δt}## - ave velocity

    3. The attempt at a solution

    a) So the first thing I did was drew a diagram of what was going on.

    Now to calculate the distance ##Δd## which is pretty straightforward, consider :

    ##Δd = d_1 + d_2 + d_3 + d_4 = 34 km##

    Now calculating the displacement ##\vec{Δd}## is a bit trickier. Looking at my diagram, I notice the first two distances allow me to form a right angle triangle with which to calculate the displacement from the origin to the end of the second distance.

    Then I think I can make another triangle from my second distance to the end of my fourth distance and calculate the displacement that way.

    After that question b) really isn't an issue. I just want to make sure my approach to the displacement is correct before I continue.
     
  2. jcsd
  3. Jun 29, 2013 #2
    What about distance 3?
     
  4. Jun 29, 2013 #3
    Yes, that will work. After that form a triangle using the displacement from distance 1 + distance 2, and distance 3 + distance 4 as 2 of the sides.
     
  5. Jun 29, 2013 #4

    HallsofIvy

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    I think the simplest way to do this is to use vector components. "14km [E]" is <14, 0>. "6 km [N]" is <0, 6>, "12 km [N 15° E] ( East 15 degrees of north )" is <12 sin(15), 12 cos(15)>, "2 km [N 65° E]" is <2 sin(65), 2 cos(64)> and add "component wise"- the displacement vector is given by <14+ 0+ 12 sin(15)+ 2 sin(65), 0+ 6+ 12 cos(15)+ 2 cos(65)>.
     
  6. Jun 29, 2013 #5

    Zondrina

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    I'm going to remember this for sure, but the thing is I'm not allowed to use this methodology to get the answer while it does look 100x more convenient.

    Okay so I formed my first triangle which is right angled. I used the pyth theorem to obtain ##c = 15.23 km##

    Then I found my angle ##θ## between ##d_1## and ##d_2## to be ##23.2°##.

    Therefore the displacement from ##d_1## to ##d_2## is ##\vec{Δd_1} = 15.23 km [E 23.2° N]##.

    Now I have one of the two displacements I need. So I formed the second triangle which is not a right angled triangle and I need to calculate the displacement from the start of ##d_3## to the end of ##d_4##.

    Using the cosine law and then the sine law should work here.

    ##c^2 = a^2 + b^2 - 2abcos(C)##
    ##c^2 = (12)^2 + (2)^2 - 2(12)(2)cos(?)##

    I'm having a bit of trouble calculating the angle between ##d_3## and ##d_4## without using dot products or advanced theorems. That should give me c. Then I simply use the sine law to calculate my angle ##\phi## for the direction.
     
  7. Jun 29, 2013 #6
    Remember that ## \vec{a} + \vec{b} = \vec{b} + \vec{a} ##. In your case that means you can sum all but the third displacement, which is easy, and then add the third displacement, so you only need to deal with angles once.
     
  8. Jun 29, 2013 #7

    Zondrina

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    This confused me a bit. Why am I adding vectors with different directions together if I'm finding displacement?

    I don't seem to see why I only need to deal with angles once when there are two triangles to consider?

    EDIT : I'm getting the angle C to be 100 degrees though I'm not positive about this.
     
    Last edited: Jun 29, 2013
  9. Jun 29, 2013 #8

    HallsofIvy

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    Because that's what displacement is- the sum of the individual vector displacements.

    You don't. You need to deal with the angles as long as you have "triangles" at all!

     
  10. Jun 29, 2013 #9
    All the other displacements are at right angles to one another, dealing with them is trivial.
     
  11. Jun 29, 2013 #10

    Zondrina

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    Alright, perhaps using the triangles isn't a good idea without a scale model. From what I'm hearing in this thread I might as well break the vectors into their respective components and figure the displacement out that way.

    Let ##\vec{Δd_E}## denote the total displacement east in km and ##\vec{Δd_N}## denote the total displacement north in km.

    Since ##d_1 = 14km [E]## we have##\vec{Δd_E} = 14##.

    Now ##d_2 = 6km [N]## so ##\vec{Δd_N} = 6##.

    ##d_3 = 12km [N 15° E]## so we need to break this vector into it's N and E components. We will have displacement 12sin(15°) [E] and 12cos(15°) [N]. Add those onto our total displacement vectors.

    Same deal for ##d_4## so after computing everything, I get ##\vec{Δd} = 26.4 km##

    Now I need to figure out the direction of ##\vec{Δd}## if I'm not mistaken. Using ##tanθ = o/a## I get ##θ = 44.23°##.

    Therefore ##\vec{Δd} = 26.4 km [E 44.23° N]##.
     
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