What does it mean for a particle to be free?

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Homework Statement


From an inertial reference frame S, the vector position of a particle of mass
m1 = 1kg is given by r1(t)=(tx-hat - t^2y-hat)m.
The vector position of a particle m2=2m1 is given by r2=(t)=(tx-hat +t^3y-hat)m

Is particle #1 free along the y-direction? Explain



Homework Equations


If the particle is free along the y-direction then the y's would equal to 0.




The Attempt at a Solution



r1= -t^2y
r2=+t^3y
=not free?
(I am actually unsure if my definition of free is true)
 
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Particle #1's motion along the y-direction is -t^2.
Would this mean that it is not free?