How to find the total distance traveled, given the position function? 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)&amp;=t^3-2t+5&amp;x(0)&amp;=5\,m&amp;x(3)&amp;=26\,m\\
v(t)&amp;=3t^2-2&amp;v(0)&amp;=-2\,m/s&amp;v(3)&amp;=25\,m/s\\
a(t)&amp;=6t&amp;a(0)&amp;=0&amp;a(3)&amp;=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.
Basically a particle will be moving in negative direction if its velocity is negative.As this type of motion is a straight line motion where $x$ is in terms of $t$ therefore total distance travelled =(distance travelled in $+v_e$ direction)+(mod of distance travelled in $-v_e$ direction)....

To solve for total distance travelled:

1.Find velocity vector by differentiating $x$ vector.

2.Find time intervals contained in the given time intervals where $v$ is $-v_e$

3.Integrate $v$ for time interval in which $v$ is $+v_e$ and add a '$-$' sign to those time time interval in which $v$ is $-v_e$ then integrate it for respective time in which $v$ is $-v_e$

4.Finally add the integrated values...