A particle moves along the x-axis obeying the equation $x = t\left( {t - 1} \right)\left( {t - 2} \right)$, where $x$ is in meter and $t$ is in second. Find the initial velocity of the particle $(m/s)$.

Hint: Initial Velocity is the velocity at time interval $t = 0$ and it is represented by $u$. It is the velocity at which the motion starts.
To determine the initial velocity of the particle, we need to expand the given equation and then differentiate it with respect to $t$.

Formula Used:
Sum or difference rule of the derivative is given by:
$(f \pm g)' = f' \pm g'$

Complete step by step solution:
In the question, the equation of the path of particle is given by $x = t\left( {t - 1} \right)\left( {t - 2} \right)$
Expand the given equation, then we have:
$x = ({t^2} - t)\left( {t - 2} \right) \\$
  $\Rightarrow x = ({t^3} - 2{t^2} - {t^2} + 2t) \\$
  $\Rightarrow x = {t^3} - 3{t^2} + 2t \\$
Differentiate the obtained equation with respect to $t$,
$v = \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}({t^3} - 3{t^2} + 2t)$
Apply the sum or difference rule to differentiate the above equation $(f \pm g)' = f' \pm g'$, then:
$v = \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {{t^3}} \right) - \dfrac{d}{{dt}}\left( {3{t^2}} \right) + \dfrac{d}{{dt}}\left( {2t} \right) \\$
 $\Rightarrow v = \dfrac{{dx}}{{dt}} = 3{t^2} - 6t + 2 \\$
And initial velocity is the velocity of particle at time $t = 0$, hence substitute $t = 0$in the value of $v$:
$v = 3{(0)^2} - 6(0) + 2 \\$
 $\Rightarrow v = 0 - 0 + 2 \\$
 $\Rightarrow v = 2m/s$

Therefore, the initial velocity of the particle is $2m/s$.

Note: As we know that the equations of motion describe a physical system's behaviour in terms of its motion. Any item subject to forces will accelerate. The object's velocity changes as a result of acceleration. Accordingly, the initial velocity is the object's speed prior to the change brought on by acceleration. The velocity will be the final velocity once the object has been accelerating for a while.