What does it mean for a particle to be free?

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The discussion centers on determining whether particle #1, with a position vector r1(t) = (tx - t^2y)m, is free along the y-direction. It is established that a particle is considered free if it is not subjected to forces, which is indicated by a constant potential or zero acceleration. The position of particle #1 shows that it has a y-component of -t^2, suggesting it experiences acceleration in the y-direction. Consequently, this indicates that particle #1 is not free along the y-direction due to its non-zero acceleration. The conclusion drawn is that particle #1 is indeed affected by forces in the y-direction.
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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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"Free" means "unaffected by forces", or equivalently that the potential is constant. In this case, it means you have to check if it's accelerating in the y-direction or not.
 
Particle #1's motion along the y-direction is -t^2.
Would this mean that it is not free?
 
That's its position at time t. What's its acceleration at time t?
 

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