Results 1 to 5 of 5

Math Help - Vector word problem

  1. #1
    Newbie Smokinoakum's Avatar
    Joined
    Mar 2011
    Posts
    23

    Vector word problem

    Hi all...I have some homework that I'm having some problems with.

    Here is the word problem that I'm confused about...

    A ship travels 18.5 km on a course of 31.2 south of east, then 12.7 km due south, then 21.5 km on the course of 61.3 west of south, where it lands. Find the displacement and the angle traveled from the starting point to the landing point.

    Here is what I did to try and solve it...

    18.5cos31.2 + 12.7cos58.8 + 21.5cos61.3 = 32.73
    18.5sin31.2 + 12.7sin58.8 + 21.5sin61.3 = 39.3

    32.73^2 + 39.3^2 * 1/2 = 51.15

    tan^-1(39.3 / 32.73) = 50.22


    so...51.15 at 50.22
    is my answer.

    For some reason, this just does not look right. Can anyone check my logic, and straighten me out here?

    Neil

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,364
    Thanks
    1311
    Where did you get the "58.8" from? It looks like you are calculating the "x" (east west) and "y" (north south) components separately. But the middle leg, 12.7 km, is "due south". There is no "east west" movement. Also, the third leg, at "61.3 west of south", since it is "west of south" should be -21.5 sin(61.3) For the "north south" movement, the middle leg is 12.7 km and the third leg 21.5 cos(61.3).

    You should have, for "x", 18.5 cos31.2 + 0 - 21.5 sin 61.3 (which is negative, so to the west) and, for "y", 18.5 sin(31.2)+ 12.7+ 21.5 cos(61.3) to the south.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie Smokinoakum's Avatar
    Joined
    Mar 2011
    Posts
    23
    Quote Originally Posted by HallsofIvy View Post
    Where did you get the "58.8" from? It looks like you are calculating the "x" (east west) and "y" (north south) components separately. But the middle leg, 12.7 km, is "due south". There is no "east west" movement. Also, the third leg, at "61.3 west of south", since it is "west of south" should be -21.5 sin(61.3) For the "north south" movement, the middle leg is 12.7 km and the third leg 21.5 cos(61.3).

    You should have, for "x", 18.5 cos31.2 + 0 - 21.5 sin 61.3 (which is negative, so to the west) and, for "y", 18.5 sin(31.2)+ 12.7+ 21.5 cos(61.3) to the south.
    Wow...I was way off in my assumptions.

    So what your saying is this...

    X = 18.5 cos(31.2) + 0 - 21.5 sin(61.3) = -3.034403707
    Y = 18.5 sin(31.2) + 12.7 + 21.5 cos(61.3) = 15.00936432

    -3.034403707^2 + 15.00936432^2 * 1/2 = 14.69943575

    Then... tan^-1(15.00936432 / -3.034403707) = -78.5707125

    I'm not sure what to do with the -78.5707125 answer for the angle. I'm guessing I need to decide what sector it's in? Can you help me a bit more?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    sure does help to make a sketch ...

    \Delta x = 18.5\cos(31.2) - 21.5\cos(28.7)

    \Delta y = -18.5\sin(31.2) - 12.7 - 21.5\sin(28.7)

    |r| = \sqrt{(\Delta x)^2 + (\Delta y)^2} \approx 32.7 \, km

    \theta = \arctan\left(\dfrac{\Delta y}{\Delta x}\right) \approx 84.7^\circ + 180^\circ or about 5.3^\circ W of S
    Attached Thumbnails Attached Thumbnails Vector word problem-vector1.jpg  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie Smokinoakum's Avatar
    Joined
    Mar 2011
    Posts
    23
    Quote Originally Posted by skeeter View Post
    sure does help to make a sketch ...

    \Delta x = 18.5\cos(31.2) - 21.5\cos(28.7)

    \Delta y = -18.5\sin(31.2) - 12.7 - 21.5\sin(28.7)

    |r| = \sqrt{(\Delta x)^2 + (\Delta y)^2} \approx 32.7 \, km

    \theta = \arctan\left(\dfrac{\Delta y}{\Delta x}\right) \approx 84.7^\circ + 180^\circ or about 5.3^\circ W of S
    Ahhhh...The picture helps a lot. Especially the angles! Thanks a LOT.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. vector word problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 12th 2011, 04:08 PM
  2. Vector word problem
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: March 22nd 2010, 05:13 PM
  3. Vector word problem
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: November 16th 2009, 01:35 PM
  4. Vector Word Problem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 19th 2009, 06:18 PM
  5. 3D Vector Word Problem Help
    Posted in the Calculus Forum
    Replies: 7
    Last Post: June 3rd 2008, 01:48 PM

Search Tags


/mathhelpforum @mathhelpforum