Suppose the acceleration of a body is given. The acceleration is second order -  {jatex options:inline}a= \frac{d^2 x}{dt^2}{/jatex}  so it might appear that we have to integrate twice and require two condition or cannot solve.{/jatex}. How can we find an expression for the velocity?
If the acceleration is given in terms of the displacement, there is a way forward. We can use the relationship  {jatex options:inline}a=\frac{dv}{dt}= \frac{dx}{dt} \frac{dv}{dx}{/jatex}  and integrate, requiring only one condition - for simultaneous values of  {jatex options:inline}v{/jatex}  and  {jatex options:inline}x{/jatex}.
Suppose a condition is  {jatex options:inline}v=1, \: x=2{/jatex}  in appropriate units.
{jatex options:inline}a=2x^2{/jatex}  and  {jatex options:inline}v=1{/jatex}  when  {jatex options:inline}tx=2{/jatex}  (in the appropriate units), then we can write
{jatex options:inline}v\frac{dv}{dx}=2x^2{/jatex}
Now separate variables.
{jatex options:inline}v dv=2x^2dx{/jatex}
Now integrate in the usual way.
{jatex options:inline}\int^v_1 vdv = \int^x_2 2x^2dx{/jatex}
{jatex options:inline}[ \frac{v^2}{2}]^v_1 = [\frac{x^3}{3}]^x_2{/jatex}
{jatex options:inline}\frac{v^2}{2}- \frac{1}{2}= \frac{x^3}{3}- \frac{8}{3}{/jatex}
Making  {jatex options:inline}v{/jatex}  the subject gives  {jatex options:inline}v= \sqrt{\frac{1}{6}(2x^3-13}{/jatex}.