## 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
$v= \sqrt{\frac{1}{6}(2x^3-13}$
.