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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?

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  • $\begingroup$ ok integrating the absolute value of the velocity: do i take the absoute value of the equation 3t^2-2? How do you do that? Just add a negative sign before it and then integrate? $\endgroup$ – user122415 Apr 14 '14 at 20:56
  • $\begingroup$ The 'net' total distance is actually displacement. You can just say you require the total distance, not the net total distance. And adding the word total to total displacement doesn't really make sense, since displacement is always a total, i.e last position - initial position. $\endgroup$ – john Jan 31 '18 at 12:55
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You should integrate the absolute value of velocity from 0 to 3. Than you get the desired result.

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    $\begingroup$ Alternatively, find all points where the velocity is $0$ and find the displacements between those points. $\endgroup$ – Alex Zorn Apr 14 '14 at 20:52
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    $\begingroup$ @user122415 i can't reply to your comment as i don't have enough reputation. you have to integrate with minus sign just before the parts, where it's negative $\endgroup$ – Tom83B Apr 16 '14 at 17:46
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if u look at the velocity function then u will find that the velocity is negative in the time interval from "0 to sq.root(2/3) sec". And it is positive in the time interval from "sq.root(2/3) to 3 sec". so take the absolute value(put an extra negative sign before velocity function) of velocity in the first time interval and integrate with in time interval b/w "0 to sq.root(2/3) sec". now again integrate velocity with in time interval b/w "sq.root(2/3) to 3 sec". now add both of the results and u will get your answer. i.e. 23.18m.

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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...

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