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Vector Addition displacement

  1. Sep 7, 2007 #1
    1. The problem statement, all variables and given/known data

    A roller coaster moves 200 ft horizontally and then rises 135 ft at an angle of 30* above the horizontal. It then travels 135 ft at an angel of 40* downward. What is its displacement from its starting point?

    2. Relevant equations

    Standard trig, I guess.

    3. The attempt at a solution

    I know that the displacement in feet is 420 ft, and I did that using trig which I could type up, but I already am 100% positive of that answer. But I need to find it in degrees too. For instance, 10* below the starting point, or 350*. But I just don't know how to get it! I drew a huge diagram on scrap paper with A - B, B - C, C - D, and A - D following the above information. Any tips or guidance would be greatly appriciated.

    Edit: I'll explain my work better. At a 30* angle, the rollercoster traveled 135 feet. Thus, the distance from the start traveled horizontally is 135cos30 =116.91 ft. Furthermore, it traveled down at a 40* angle, which is 50* in respect to the ground, so 135sin50=103.4 ft. 200 ft + 117 ft + 103 ft = 420 ft, which is the answer I got and what the "webassign" (computer program I use in class for homework) got (it tells you if the answer is right). Now I need to figure out how many degrees below the horizontal (starting point) the rollercoster ended up. Any help from this point is appreciated greatly ^^
    Last edited: Sep 7, 2007
  2. jcsd
  3. Sep 7, 2007 #2


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    420 ft (rounded to the nearest foot) is the horizontal component of the displacement. What is the vertical component?

    When you have both the horizontal and vertical components, you should be able to compute the magnitude and direction (in degrees above or below the horizontal axis, depending on whether the vertical component of displacement turns out positive or negative) of the displacement.

    (I assumed that "rises 135 ft at an angle of 30*" means it travels a distance of 135 ft, not that it's change in height is 135 ft. It's not clear that this is the right interpretation.)
    Last edited: Sep 7, 2007
  4. Sep 7, 2007 #3
    Ah.. so the vertical component is what I'm missing. That would certainly make things much easier. Is the vertical component how much the roller coster travels from the end of the 200 feet to the end of its decline?
  5. Sep 7, 2007 #4
    You're right though, it travels 135 feet but does not gain 135 feet in height above the ground. It goes up at a 30* angle and moves a total of 135 feet
  6. Sep 7, 2007 #5
    well the total displacement rounded to 2 significant figures is 420 o.o

    But yea get the x/y components of the 3 vectors added up and you'll get the angle that it makes.

    In order to get the components of vectors you do this:
    if you have vector z with equation z=r+theta.
    R is the maginute and theta is the angle which can be + or - depending on the question. Your problem says 40* but 40* up? no down so it's actually -40* for that vector.

    x-component of z = rcos(theta)
    y-component of z = rsin(theta)

    add up your 3 vector's 2 components each.

    total vector say...S

    x component of S = x components of other vectors added together.
    y compoment of S = y components of other vectors added together.


    [tex]\tan\theta = \frac{S_y}{S_x}[/tex]

    Edit: ok yea you got the x components now do the same for the y components. Since you know tan (theta) = the sum of components y/x so then theta = ?
    Last edited: Sep 7, 2007
  7. Sep 7, 2007 #6
    So for the y
    135sin30 = 67.5
    135cos50 = 86.78
    So I get together 154 (rounded) for the y component
    and thetha would equal around 20*
    So I would end up getting 20*... below the horizontal? In other words, 340* counterclockwise?
  8. Sep 7, 2007 #7
    Or would it be above? On fruther notice, I saw no negatives in my calculator, so it is 20* above the horizontal?
  9. Sep 7, 2007 #8
    hm..ok wait you did a few things wrong but still got the right answer xD

    in your x components all are r *cos (theta)

    40* down is -40* with respect to the x-axis
    so it should be 135cos(-40) instead of 135sin(50) but the answer is the exact same thing (420)

    when you solve for the y components:
    the first one's right but for the second one use sin and -40* which it should be and post what you get (I got some smalll angle)
  10. Sep 7, 2007 #9
    135sin(-40) = -86.776
    then adding that to 67.5 you get -19.2763
    the degree would be -2.627, which I would round to 3
    So the answer is 3* below the horizontal or 357* counterclockwise?
  11. Sep 7, 2007 #10
    yea -3* (357*) or maybe -2.63* (357.37*) depending on how many significant figure's the website asks for.

    edit: no problem glad I could help ya out
    Last edited: Sep 7, 2007
  12. Sep 7, 2007 #11
    Thank you very much! I can honestly say I understand what I did wrong and this will greatly help me! I'm glad I didn't give up on the problem. Thanks so much again! ^_^
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