The acceleration of a particle is increasing linearly with time $t$ as \[bt\]. The particle starts from the origin with an initial velocity ${V_0}$. The distance travelled by the particle in time $t$ will be _____?

Hint: If the acceleration of the particle changes with the flow of time, in that case the acceleration of the particle is a complex motion and the particle's acceleration is difficult to analyze. But if the acceleration of a particle is increasing linearly with time then the relation of particle's velocity, acceleration and displacement would be easy to explain numerically.

Formula Used:
\[a = \dfrac{v}{t}\]
$v = \dfrac{s}{t}$
Where,
$a$: acceleration, $v$: velocity, $s$: displacement, $t$: time.
Formula of Integration:- $\int {{x^n}} dx$
$ = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$
Where $c$ is constant of integration, whose value can be anything.

Complete step by step answer:
$a = \dfrac{v}{t} = \dfrac{{dv}}{{dt}}$
Because the acceleration of a particle is increasing linearly with time, so:-
$a = bt$
Integrate this and we get $v$,
$\int {bt} $
$\Rightarrow$ $ = \int {\dfrac{{b{t^{1 + 1}}}}{{1 + 1}}} $
$\Rightarrow$ $v = \dfrac{{b{t^2}}}{2} + c$
The particle starts from the origin with an initial velocity ${V_0}$, so:-
${V_0} = c$
$\Rightarrow$ $v = \dfrac{{b{t^2}}}{2} + {V_0}$
$\Rightarrow$ $v = \dfrac{s}{t} = \dfrac{{ds}}{{dt}}$
$\dfrac{{ds}}{{dt}} = \dfrac{{b{t^2}}}{2} + {V_0}$
$\Rightarrow$ $ds = (\dfrac{{b{t^2}}}{2} + {V_0})dt$
Integrating the equation we get:-
$\Rightarrow$ $ds = (\dfrac{1}{2}\int {\dfrac{{b{t^{2 + 1}}}}{{2 + 1}} + \int {{V_0}} )} $
$\Rightarrow$ $s = (\dfrac{1}{6}b{t^3} + {V_0}t) + c$ and $c = 0$
Hence, $s = \dfrac{{b{t^3}}}{6} + {V_0}t$

Note: When the particle changes his velocity in between time-period is called acceleration of the particle. Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity. An object is accelerating if it is changing its velocity.