Kinematic Equation of Motion for uniformly accelerated motion - Calculus method
(i) Velocity attained by a particle after time t.
Let dv be the change in velocity of the particle in time dt . Therefore, the acceleration of the particle is given by
\[a= \frac{dv}{dt}\] \[\ dv = a dt\]
Intergrating both sides
\[\int_{v_{0}}^{v}dv=\int_{0}^{t}adt=a\int_{0}^{t}dt\] \[v-v_{0}=at\]
(ii) Displacement of the Particle after time t.
\[v=\frac{dx}{dt}\] \[dx=vdt\]\[intergrting \;both\;sides\]\[\int_{x_{0}}^{x}dx = \int_{0}^{t}vdt\]
\[=\int_{0}^{t}(v_{0}+at)dt\] \[x-x_{0}=v_{0}t+\frac{1}{2}at^{2}\]\[x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\]
(iii) Velocity attained by a particle after travelling a distance S :
\[a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}\]\[or,vdv=adx\]
Intergrating both sides,
\[\int_{v_{0}}^{v}vdv=\int_{x_{0}}^{x}adx\]\[\frac{v^{2}-v_{0}^{2}}{2} = a(x-x_{0})\]\[v^{2}=v_{0}^{2}+2(x-x_{0})\]