Answer By definition

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\mathrm{d} v=a \mathrm{d} t$

Integrating both sides

$\int_{u_{0}}^{v} \mathrm{d} v=\int_{0}^{t} a d t$

$=a \int_{0}^{t} \mathrm{d} t \quad \text { (a is constant) }$

Further,

$\boldsymbol{v}-\boldsymbol{v}_{0} =\boldsymbol{a} t$

$\boldsymbol{v} =\boldsymbol{v}_{0}+\boldsymbol{a} t$

$\boldsymbol{v} =\frac{\mathrm{d} \boldsymbol{x}}{\mathrm{d} t}$

$\mathrm{d} x=v \mathrm{d} t$

Integrating both sides

$\int_{x_{0}}^{x} \mathrm{d} x=\int_{0}^{t} v \mathrm{d} t$

$=\int_{0}^{t}\left(v_{0}+a t\right) \mathrm{d} t$

$x-x_{0} =v_{0} t+\frac{1}{2} a t^{2}$

$x =x_{0}+v_{0} t+\frac{1}{2} a t^{2}$

We can write

$a=\frac{d v}{d t}=\frac{d v}{d x} \frac{d x}{d t}=v \frac{d v}{d x}$

$\text { or, } v \mathrm{d} v=a \mathrm{d} x$

ntegrating both sides,

$\int_{u_{0}}^{v} v \mathrm{d} v=\int_{x_{0}}^{x} a \mathrm{d} x$

$\frac{v^{2}-v_{0}^{2}}{2}=a\left(x-x_{0}\right)$

$v^{2}=v_{0}^{2}+2 a\left(x-x_{0}\right)$