Answer:
Explanation:
We can prove that this is a circular motion if we demonstrate that the norm of the vector is independent of time. Hence we have
[tex]R(t)=2.0cos(3t)\hat{i}+2sin(3t)\hat j\\\\|R(t)|=\sqrt{4cos^2(3t)+4sin^2(3t)}\\\\|R(t)|=\sqrt{4(cos^2(3t)+sin^2(3t))}\\\\[/tex]
but cos^2(3t)+sin^2(3t)=1. Thus we obtain
[tex]|R(t)|=\sqrt{4}=2[/tex]
The norm is independent of t, thus, the particle describes a circular motion
Hope this helps!!