# Vectors and finding relative speed

Tags:
1. Oct 13, 2014

### bluetriangle

1. The problem statement, all variables and given/known data
Find Particle A's speed with respect to Particle B.

I am given:
Particle A: (15i - 10j) or 18.03 m/s
angle A: -33.69 degrees

Particle B: (5i + 15j) or 15.81 m/s
angle B: 71.57 degrees

(Both angles with respect to x axis)

2. Relevant equations
Had there not been any angles, I know I would have to subtract the two vector quantities but I don't know how to solve it with angles.

3. The attempt at a solution
I know the k components, found by using Pythagorean Thereom, do I simply subtract k(A) - k(B) ?

that would be 37.88 k.
Is this correct?

2. Oct 13, 2014

### BvU

Hi Blue, and welcome to PF :-)

"Had there not been any angles": well, simply ignore them and use the vector components...

Under 2. Relevant equations, I don't see even a single equation.

Under 3. Attempt at solution, you demonstrate a severe lack of understanding the problem statement.

The $\bf \hat\imath$ and $\bf \hat\jmath$ are unit vectors in the x- and y direction, respectively. That should tell you the velocity vectors are in the x-y plane and the z component is zero. I have no idea what you think you are doing when you calculate k(A) or k(B). You don't show your work either, so it is very difficult for me to guess how you manage to come up with 37.88 k. Where does it come from ? What does it mean ? Particle A's speed with respect to particle B is 37.88 k ?

There is one way I can find 37.875, but i shudder to type it out.

If the exercise as given is too complicated, try an easier one to start with: same question for Particle A: (15i - 10j) Paricle B: (0i + 0j)