1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

One dimension kinematics, what am I doing wrong?

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Car A is located 43km away, 17 degrees north of west. Car B is 34km away, 38 degrees east of north. The person in car A checks the position of car B by using a GPS. What does it give for car B's
    a) distance from car 1
    b) direction, measured due east?

    2. Relevant equations
    Pythagorean Theorem
    Using sin, cos, tan


    3. The attempt at a solution
    Ax = (cos17)(43) = 41.12km
    Ay = (sin17)(43) = 12.57km
    Bx = (cos38)(43) = 26.79
    By = (sin38)(34) = 20.93km

    Cy = By - Ay
    Cy = 20.93km - 12.57km
    Cy = 8.36km

    Cx = Bx + Ax
    Cx = 26.79 + 41.12km
    Cx = 67.91km

    Vector C = sqrt[(Cy^2 + Cx^2)]
    Vector C = sqrt[(8.36^2 + 67.91^2)]
    a) Vector C = 68.42km

    Angle = tan(Cx/Cy)
    Angle = 67.91/8.36
    b) Angle = 82.48 degrees
     
  2. jcsd
  3. Sep 21, 2013 #2

    PeterO

    User Avatar
    Homework Helper

    You have the components of B back to front.

    It is not good enough to just use cos for the x component and sin for the y component.
    The 38 angle is from the vertical, North, where as the 34 angle was from the horizontal, West.
     
  4. Sep 21, 2013 #3
    I don't get it what you're trying to say
     
  5. Sep 21, 2013 #4

    PeterO

    User Avatar
    Homework Helper

    Draw a sketch of the situation - making sure that your sketch versions of 34 degrees and 38 degress are actually smaller than 45 degrees.

    For 34 degrees North of west, it will be obvious that the x component is bigger than the y component.

    For the 38 degrees east of North, it should be equally obvious that the x component is smaller that the y component.

    But in your calculations, you have Bx > By.

    Alternatively

    In the triangle formed for the "34 North of West" location, the y component is opposite the angle - so use sine, as you did - while the x component is adjacent to the angle - so use cosine, as you did.
    In the triangle formed for the "38 East of North" location, the y component is adjacent to the angle - so use cosine, but you used sine - while the x component is opposite the angle - so use sine, but you used cosine.
     
  6. Sep 21, 2013 #5
    Oh east of north, so what i did in my calculation was north of east?
    would i still get a right answer if did 90-38 = 52 degrees to have the y component opposite to the angle
     
  7. Sep 21, 2013 #6

    PeterO

    User Avatar
    Homework Helper

    Yes you would - but why would you bother?

    Why your strong desire for the y component to be a sine - eventually (like here) you are going to get it backwards and get it wrong.

    My approach is to work out both components - you often need them both anyway - then do a quick sketch - ensuring that if an angle is less than 45 degrees make sure it is sketched as such (better to make it too small than too big.

    From the sketch you can see whether you want the smaller or larger component, and so just select the smaller or larger one as appropriate.

    ie. don't spend time deciding whether you should be using sin or cos, concentrate on whether you want the small component or the large component.

    example: a projectile is fired at 200m/s at an angle of 60 degrees to the horizontal - what is the vertical component of its initial velocity.
    ans: the two components are 200.sin60 and 200.cos60 or 173.2m/s and 100 m/s
    A quick sketch (perhaps just an imagined sketch) shows that the projectile is travelling faster up than horizontally, so the vertical component must be 173.2 m/s

    compare that to: a cannonball is fired at 200m/s at an angle of 60 degrees to the vertical - what is the vertical component of its initial velocity.
    ans: the two components are 200.sin60 and 200.cos60 or 173.2m/s and 100 m/s
    A quick sketch (perhaps just an imagined sketch) shows that the projectile is travelling faster horizontally than vertically, so the vertical component must be 100 m/s
     
  8. Sep 21, 2013 #7

    PeterO

    User Avatar
    Homework Helper


    Note: when drawing the sketches, the key is whether the angles involved are smaller or larger than 45 degrees (at 45 degrees the components are equal)

    In my sketches, I would draw both 34 and 38 degrees very small - like a real 10-15 degrees to exaggerate, for me, that one component is smaller than the other. The last thing you want is for you sketch, with its wonky lines, showing the components approximately the same size.
     
  9. Sep 21, 2013 #8
    Ah okay well thank you very much! what got me confused was the directions, such as "north of east" or "east of north" and thats what mixed me up! i got the right answer now and well anyways thank you so much i appreciate you helping me out. the sketches definitely help out A LOT
    Anyways I pressed the "thanks" button thank you so so much
     
  10. Sep 21, 2013 #9

    PeterO

    User Avatar
    Homework Helper

    happy to help - and good to see you (now?) see the value of a simple sketch.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: One dimension kinematics, what am I doing wrong?
  1. What am I doing wrong? (Replies: 15)

  2. What Am I Doing Wrong? (Replies: 2)

  3. What am I doing wrong? (Replies: 1)

Loading...