Question

a student walks 50m on a bearing 0.25 degrees and then 200m due east how far is she from her starting point.​

Answer 1

Answer:

Step-by-step explanation:

I'm going to use Physics here for this concept of vectors. Here are some stipulations I have set for the problem (aka rules I set and then followed throughout the problem):

** I am counting the 50 m as 2 significant digits even though it is only 1, and I am counting 200 as 3 significant digits even though it is only 1. 1 sig dig doesn't really give us enough accuracy, in my opinion.

** A bearing of .25 degrees is measured from the North and goes clockwise; that means that measured from the x axis, the angle is 89.75 degrees. This is the angle that is used in place of the bearing of .25 degrees.

** Due east has an angle measure of 0 degrees

Now let's begin.

We need to find the x and y components of both of these vectors. I am going to call the first vector A and the second B, while the resultant vector will be C. Starting with the x components of A and B:

$A_x=50cos(89.75)$ so

$A_x=.22$

$B_x=200cos(0)$ so

$B_x=200$ and we need to add those results together. Due to the rules for adding significant digits properly, the answer is

$C_x=200$ (and remember I am counting that as 3 sig fig's even though it's only 1).

Now for the y components:

$A_y=50sin(89.75)$ so

$A_y=50$ (which I'm counting as 2 sig fig's)

$B_y=200sin(0)$ so

$B_y=0$ and we need to add those results together.

$C_y=50$

Now for the resultant magnitude:

$C_{mag}=\sqrt{(200)^2+(50)^2}$  and that gives us a final magnitude of

$C_{mag}=206$ m

Now for the angle:

Since both the x and y components of the resultant vector are in quadrant 1, we don't need to add anything to the angle to get it right, so

$tan^{-1}(\frac{50}{200})=14$

The girl is 206 meters from her starting point at an angle of 14 degrees