Given, position of the particle is x=4−12t+3t2 Differentiating both sides with respect to time (t), we get
dx
dt
=0−12+3(2t)=6t−12 ∴v(t)=6t−12 Att=1s,v=6×1−12=−6m∕s At t=0s,x=4m At t=1s,x=4−(12×1)+(3×12)=−5m Hence, particle is moving along negative x direction.