Solving for Acceleration In Terms of Displacement

Suppose the acceleration of a body is given. The acceleration is second order -  
\[a= \frac{d^2 x}{dt^2}\]
  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  
\[a=\frac{dv}{dt}= \frac{dx}{dt} \frac{dv}{dx}\]
  and integrate, requiring only one condition - for simultaneous values of  
\[v\]
  and  
\[x\]
.
Suppose a condition is  
\[v=1, \: x=2\]
  in appropriate units.
\[a=2x^2\]
  and  
\[v=1\]
  when  
\[tx=2\]
  (in the appropriate units), then we can write
\[v\frac{dv}{dx}=2x^2\]

Now separate variables.
\[v dv=2x^2dx\]

Now integrate in the usual way.
\[\int^v_1 vdv = \int^x_2 2x^2dx\]

\[[ \frac{v^2}{2}]^v_1 = [\frac{x^3}{3}]^x_2\]

\[\frac{v^2}{2}- \frac{1}{2}= \frac{x^3}{3}- \frac{8}{3}\]

Making  
\[v\]
  the subject gives  
\[v= \sqrt{\frac{1}{6}(2x^3-13}\]
.

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