A particle moves along the x-axis such that its position at time t is given by $x(t)=2t^3-6t^2+4t$. Find the total distance traveled by the particle in the interval $t=0$ to $t=3$ seconds.

Solution

1. The velocity function is $$\begin{array}{rl}v(t)& =\frac{dx}{dt}\\ & =6t^2-12t+4\end{array}$$.
2. Setting $v(t)=0$ gives $t=1\pm \frac{\sqrt{3}}{3}$, which are the times when the particle changes direction.
3. Calculate the positions at the endpoints and turning points: $x(0)=0$, $x(1-\frac{\sqrt{3}}{3})=\frac{2\sqrt{3}}{9}+2$, $x(1+\frac{\sqrt{3}}{3})=2-\frac{2\sqrt{3}}{9}$, and $x(3)=12$.
4. The total distance is $$\begin{array}{rl}|x(1-\frac{\sqrt{3}}{3})-x(0)|+|x(1+\frac{\sqrt{3}}{3})-x(1-\frac{\sqrt{3}}{3})|+|x(3)-x(1+\frac{\sqrt{3}}{3})|& =(2+\frac{2\sqrt{3}}{9})+\frac{4\sqrt{3}}{9}+(10+\frac{2\sqrt{3}}{9})\\ & =12\end{array}$$.
5. Hence the answer is (C).