Originally Posted by

**Soroban** Hello, metallica007!

Who uses the term "placement"?

The location of the particle is usually called its Position.

The position function is the integral of the velocity function.

$\displaystyle s(t) \;=\;\int (t^2 - t)\,dt \;=\;\tfrac{1}{3}t^3 - \tfrac{1}{2}t^2 + s_o $

. . where $\displaystyle s_o$ is the particle's initial position (at $\displaystyle t = 0$).

At $\displaystyle t=0\!:\;\;s(0) \:=\:\tfrac{1}{3}(0^3) - \tfrac{1}{2}(0^2) + s_o \;=\;s_o$

At $\displaystyle t = 5\!:\;\;s(5) \;=\;\tfrac{1}{3}(5^3) - \tfrac{1}{2}(5^2) + s_o \;=\;\frac{175}{6} + s_o$

During that time interval, its displacement is:

. . $\displaystyle s(5) - s(0) \;=\;\left(\frac{175}{6}+s_o\right) - s_o \;=\;\frac{175}{6} \;=\;29\tfrac{1}{6}\text{ meters}$

The particle changes direction when $\displaystyle v = 0.$

. . $\displaystyle t^2-t \:=\:0 \quad\Rightarrow\quad t(t-1) \:=\:0 \quad\Rightarrow\quad t \;=\;0,\;1$

At $\displaystyle t=0$, its position is: .$\displaystyle s(0) = s_o$

At $\displaystyle t = 1$, its position is: .$\displaystyle s(1) \:=\:\tfrac{1}{3}(1^3) - \tfrac{1}{2}(1^2) + s_o \;=\;s_o-\tfrac{1}{6}$

Its displacement is: .$\displaystyle s(1) - s(0) \;=\;\left(s_o - \tfrac{1}{6}\right) - s_o \;=\;-\tfrac{1}{6}$

. . In the first second, the particle moved $\displaystyle \tfrac{1}{6}$ meter *to the left.*

At $\displaystyle t = 5$, its position is: .$\displaystyle s(5) \;=\;\tfrac{1}{3}(5^3) - \tfrac{1}{2}(5^2) + s_o \;=\;\tfrac{175}{6} +s_o$

Its displacement is: .$\displaystyle s(5) - s(1) \;=\;\left(\tfrac{175}{6} + s_o\right) - \left(s_o - \tfrac{1}{6}\right) \;=\;\tfrac{88}{3}$

. . In the next 4 seconds, it moved $\displaystyle 29\tfrac{1}{3}$ meters *to the right.*

Therefore, it moved a total distance of: .$\displaystyle \tfrac{1}{6} + 29\tfrac{1}{3} \;=\;29\tfrac{1}{2}$ meters.