Vector Problem involving Cartesian plane

AI Thread Summary
The discussion revolves around a physics problem involving two particles with different accelerations on the Cartesian plane. Particle 1 accelerates at 5.00 m/s² along the x-axis, while Particle 2 accelerates at -6.00 m/s² along the y-axis, both starting from rest. The main question is whether to calculate the resultant speed of Particle 1 with respect to Particle 2 after 4 seconds, which is found to be approximately 31.2 m/s using the Pythagorean theorem. The conversation also explores an alternative method involving derivatives to find the rate of change of the resultant distance. Clarification on the problem's wording is sought to ensure the correct interpretation of the required calculation.
proster
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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!
 
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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.
 
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.
 
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