Vector problem

[gogogadget]

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.

Relevant equations for vectors:

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

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!

 

[srmichael]

Hi gogogadget.

Unless you are specifically being told to use Law of Cosines, I approach vector problems in a different way, a way that I think is less confusing.

Remember that any vector can be written as a coordinate like this: (AcosΘ,AsinΘ) where A is the magnitude of the object and Θ is the directional angle. So for the car travelling 20km north, that vector can be written as:

(20cos90°, 20sin90°)

Then when the car drives northwest at 60 degrees, that vector can be written as:

(35cos150°, 35sin150°) (do you see why I it's 150°?)

Then, simply calculate each of these coordinates, add them up, and then that resulting coordinate is the coordinate of the resulting displacement vector.

(20cos90°, 20sin90°) = (0, 20)
(35cos150°, 35sin150°) = (-30.31, 17.50)

Displacement vector (x, y) = (-30.31, 37.50)

With this, we can calculate the magnitude which is \(\displaystyle \sqrt{x^2+y^2}\). Thus:

Magnitude = \(\displaystyle \sqrt{(-30.31)^2+(37.50)^2}=48.2km\)

The resulting directional angle is then \(\displaystyle \tan^{-1}(\frac{y}{x})+\pi\). Adding \(\displaystyle \pi\) is only necessary when x < 0 in order to put us in the proper quadrant. Thus:

Direction = \(\displaystyle \tan^{-1}(\frac{37.50}{-30.31})+\pi=2.25\ radians=128.9°\)