Kinematic Equation of Motion for uniformly accelerated motion - Calculus method

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,

\[\int_{v_{0}}^{v}vdv=\int_{x_{0}}^{x}adx\]\[\frac{v^{2}-v_{0}^{2}}{2} = a(x-x_{0})\]\[v^{2}=v_{0}^{2}+2(x-x_{0})\]