# Relative Motion

1. Jul 7, 2013

### jwroblewski44

Disclaimer: This is not for any class. I have downloaded this book ( http://www.ck12.org/book/CK-12-Physics---Intermediate/ ) and am ( attempting to (: ) work through it on my own. I am going through the section on relative motion ( section 3.3 in the book ) and am confused by a portion. It goes: (copy/pasted from book)

Car A is moving due east with a speed of 30 mph and Car B is moving due north with a speed of 30 mph.
a. What is the velocity of car A relative to car B?
Vab =~Va - Vb The two vectors are not along the same line so we’ll use their components
Va = (+30;0) and Vb = (0;+30), where east is +x and north is +y
-Vb = (0;-30), therefore, Va - Vb = (30+0;0-30) = (+30;-30). The components are directed east and south, so
the direction is southeast. Since both components have the same magnitude, the angle must be 45. But since the
vector is in the southeast direction it is in the 4th quadrant so the angle is 315, and the magnitude is the Pythagorean
sum (30[^]2 +30[^]2) [^]1/2 = 42:4 mph.

What is most confusing is when they state, "Since both components have the same magnitude, the angle must be 45." I see that both cars (vectors?) have the same magnitude (30mph?), but how does that mean "the angle must be 45"? And then they state that because the vector has a southeastern direction it is in the 4th quadrant, so the angle is 315... Where does the angle of 315 come into play at all? I see them state the angle then not use it directly in any formula.

Can someone help shine some light on this matter for me? Thanks in advance!

EDIT: It seems I have posted in the wrong area. Sorry :)

Last edited: Jul 7, 2013
2. Jul 7, 2013

### DrewD

First, the vectors that are being referred to are the velocity vectors, which is different from an actual position vector. I couldn't tell if you were clear about this or not. Since the components are perpendicular to each other, the triangle that these velocity components make is a right triangle with the two smaller sides of equal length. Since the other angles are equal (there are a number of ways to show this), and the sum of the internal angles of a triangle equal 180$^{\circ}$, the two smaller angles must both be 45$^{\circ}$

Now, I'm assuming there is a picture that shows the two cars starting from the same point and travelling away from each other. From this picture and the other work that has been done, we see that the velocity vector is pointing 45$^{\circ}$ south of east. This is the same as measuring 315$^{\circ}$ anti-clockwise from east. Since this is a commonly used orientation, they ask you to put it in that form.

Does that clear it up? This stuff can be a bit annoying, but the geometry practice is very important for other topics.

I think that these should usually be in the classwork help even though it isn't for a class, but I'm not sure about that.

3. Jul 8, 2013

### jwroblewski44

Thanks for the explanation! Pointing out the reference to a right triangle helped me to understand.

Last edited: Jul 8, 2013