b) The particle changes direction when its velocity changes sign. Use the first derivative test to find the critical points:
[tex]v(t)=2(t-3)^2(2t-3)=0\implies t=\dfrac32\,\text{ or }t=3[/tex]
Then for [tex]t<\dfrac32[/tex], we have [tex]v(t)<0[/tex]; for [tex]\dfrac32<t<3[/tex], we have [tex]v(t)>0[/tex]; and for [tex]t>3[/tex], we have [tex]v(t)>0[/tex] again. These three facts indicate that the velocity only changes once at [tex]t=\dfrac32[/tex].
c) Pick "to the right" to be the positive direction. Then you need to find where [tex]v(t)>0[/tex]. We already did that above and found [tex]v(t)>0[/tex] for [tex]t\in\left(\dfrac32,3\right)\cup(3,\infty)[/tex].
