## Solving for Acceleration In Terms of Displacement

\[a= \frac{d^2 x}{dt^2}\]

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 \[v= \sqrt{\frac{1}{6}(2x^3-13}\]

.