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