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Vector car problem

  1. Dec 18, 2012 #1
    1. The problem statement, all variables and given/known data

    I was going to see if anyone might be able to provide with clarification on how to go about solving this problem? It was done in my book as an example but I didn't follow the logic to solving it.


    A car travels 20.0 km due north and then 35.0 km in a direction 60 degrees northwest as shown in the attached document. Find the magnitude and direction of the car's resultant displacement.



    2. Relevant equations

    Relevant equations for vectors:

    x = r cos Θ
    y = r sin Θ
    r = √x2 + y2
    tan Θ = y/x

    3. The attempt at a solution



    In the problem, they use the law of cosines to find R (the magnitiude of the car's displacement) from the equation:

    R2 = A2 + B2 - 2AB cos Θ

    To find Θ, they take:

    Θ = 180° - 60° = 120°

    From there, they set up the equation as:

    R = √AA2 + B2 - 2AB cos Θ

    In substituting the numbers given in the problem, they find a final answer of 48.2 km. For the equation above, I understand that the pythagorean theorem is applied for part of it and that they take the equation for x as we are measuring for displacement in the x direction and using the vectors of r as shown in the document. My question for this part, why do they find -2AB for it? Is it because there are two vectors?

    For the direction, they solve for R in the northerly direction with angle beta. But they set it up as:

    sin β/β = sin Θ/R

    I don't follow the setup for it. I'm not sure why it's all set up as a fraction for the equation? Any input would be appreciated!
     

    Attached Files:

  2. jcsd
  3. Dec 18, 2012 #2

    PhanthomJay

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    Homework Helper
    Gold Member

    There are several ways to calculate the resultant of two vectors, and using the law of cosines to find the magnitude of the resultant is one of them, which then requires using the law of sines to find the angle (direction) of the resultant (which you typed wrong , should be sin β/B = sin Θ/R).
    It is often easier to find the resultant by breaking up each vector into its x and y components, and then algebraically add all the x components together to get R_x, add all the y components together to get R_y, and magnitude of R is the sq rt of the sum of those squares (per Pythagoras), and direction of R is the angle whose tangent is R_y/R_x. Always draw a rough sketch to approximate scale, and watch plus and minus signs.
     
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