A particle moves in a straight line and is at a distance $x$ from a fixed point $O$ on the line at time $t$. If $$12x = 27 + 6t^2 - t^4$$ find the velocity and acceleration of the particle at time $t$. Express the acceleration in terms of $x$ and prove that $x$ never exceeds $3$.
My attempt ...
$x = \frac{27}{12} + \frac{t^2}{2} - \frac{t^4}{12}$
$v ={\frac{dx}{dt} = t - \frac{t^3}{3}}$
$a = \frac{dv}{dt} = 1 - t^2$
The above was easy. How to express the acceleration in terms of $x$ and prove that $x$ never exceeds $3$
I was thinking of solving the equation for $t^2$ and subtituting in the equation for acceleration above $$t^4 -6t^2 + (12x -27) = 0$$
$$t^2 = \frac{6 \pm \sqrt {36 - 4(12x - 27)}}{2}$$ $$t^2 = 3\pm2\sqrt 3\sqrt{(3-x)}$$ $$a = -2\pm2\sqrt 3\sqrt{(3-x)}$$
Not sure if there is a better approach to this problem than this. However, I am not sure how to prove that $x$ never exceeds $3$.
Is it just to keep the term inside the square root positive?
$$(3 - x) \ge 0$$ $$x \le 3$$