Vector Problem involving Cartesian plane

[proster]

My main issue is trying to understand the question being asked (is it asking for the magnitude of the resultant vector for speed?)

Homework Statement

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.

Homework 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.

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!

 

[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.

 

[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.