Physics - Kinematics Question with Solution | TestHub

The velocity is $v(t)=\frac{dx}{dt}=3t^2-12t+9=3(t-1)(t-3)$. The particle changes direction when $v(t)=0$, i.e., at $t=1\text{ s}$ and $t=3\text{ s}$. Position at key times:$x(0)=0$, $x(1)=1-6+9=4\text{ m}$, $x(3)=27-54+27=0\text{ m}$, $x(4)=64-96+36=4\text{ m}$. Total distance covered$=|x(1)-x(0)|+|x(3)-x(1)|+|x(4)-x(3)|=|4-0|+|0-4|+|4-0|=4+4+4=12\text{ m}$.