# Vector Problem involving Cartesian plane

1. Sep 5, 2010

### proster

My main issue is trying to understand the question being asked (is it asking for the magnitude of the resultant vector for speed?)
1. The problem statement, all variables and given/known data
Particle 1 is moving on the x-axis with an
acceleration of 5.00 m/s^2 in the positive x-
direction. Particle 2 is moving on the y-
axis with an acceleration of 6.00 m/s^2 in the
negative y-direction. Both particles were at
rest at the origin at t = 0 s.
Find the speed of particle 1 with respect to particle 2 at 4.00 s. Answer in m/s.

2. Relevant equations
I know that this is a problem involving constant acceleration so:
$$v = v_0 + a t$$ One for the x direction and one for the y direction.

3. The attempt at a solution
So V0x and V0y are both 0. Then, Vxf=(5m/s^2)(4s)=20 m/s and Vyf=(-6m/s^2)(4s)=-24 m/s. Now the problem is whether the problem wants me to find the resultant speed (the wording is somewhat confusing for me). If so, then sqrt(20^2+(-24)^2)) or about 31.2 m/s. If not, I have no idea. I'd appreciate if someone could let me know if I'm on the right track. Thanks!

2. Sep 6, 2010

### rl.bhat

Hi proster, welcome to PF.

If x is the distance traveled by particle 1 along x-axis and y is the distance traveled by the particle 2 along the y-axis, then

x^2 + y^2 = R^2.

By taking the derivative you get

2x*dx/dt + 2y*dy/dt = 2R*dR/dt..........(1)

x = 1/2*a(x)t^2 and dx/dt = V(x) = at.

y = 1/2*a(y)t^2 and dy/dt = V(y) = at.

R^2 = x^2 + y^2 So R = sqrt(x^2 + y^2)

Substitute these values in eq.(1) and find dR/dt.

3. Sep 6, 2010

### proster

Thanks a lot! I got the same answer as the way I did it the first time, but it's definitely an interesting way of thinking about the problem.