A particle moves in a straight line according to the rule $x(t)=t^3-2t+5$, where $x(t)$ is given in meters and where $t$ is given in seconds. Determine the position, velocity, and acceleration of the particle at $t=0$ and $t=3$ seconds. How far has the particle moved during this $3$ second period?
Answer
\begin{align*}x(t)&=t^3-2t+5&x(0)&=5\,m&x(3)&=26\,m\\ v(t)&=3t^2-2&v(0)&=-2\,m/s&v(3)&=25\,m/s\\ a(t)&=6t&a(0)&=0&a(3)&=18\,m/s^2\end{align*}
Total distance traveled is $23.18$m.
My question concerns the total distance traveled. I know by definition distance is the total displacement (the net total distance, regardless of direction). But how do you get $23.18$ m from the equations?
You should integrate the absolute value of velocity from 0 to 3. Than you get the desired result.