I only need to know how to calculate time (t), I'll be able to do everything else after that by myself, and I've really tried, but I reach a hellish integral (that I've tried to calculate, but my teacher told me that it wasn't the way, his advice was that I should use the initial conditions in a clever way).
I'll upload a picture with clear calculations with all that makes sense. The problem is as follows:
Knowing that a point mass has an acceleration a(x) = -x - 3x² , find time (t), position ( x(t) ) and velocity ( v(t) ). Initial conditions are:
v = 2 m/s when x = -1 m , t = 2 s
A picture of my progress ---> Exercice
I'm not supposed to integrate like a mad man, instead, the teacher told me that there's a clever way to solve v(x) to find time t, but it requires some finesse with calculus and curves.
Many thanks, don't be afraid to throw advanced concepts in order to solve it, I do know differential equations and advanced mechanics. This is a one dimmensional problem from a 1st (introductory) class of physics. Not even Goldsteins mechanics kicked me that hard XD. Thanks for the help.
$${\ddot{x}= -x - 3x²} \\ v\frac{dv}{dx}=-x - 3x² \\ \int_2^v v{dv}=\int_{-1}^x-x - 3x²\ dx \\ \frac{v^2}{2}-2=\frac{-x^2}{2}-x^3+\frac{1}{2}-1 \\ {v^2}={-x^2}-2x^3+{3} \\ {v}=\sqrt{{-x^2}-2x^3+{3}}\\ \int_{-1}^x\frac{dx}{\sqrt{{-x^2}-2x^3+{3}}}=\int_2^tdt$$
This looks not so good but good news is that, the term $2x^3$ has been integrated in the interval of $(-1,x)$, here $x$ never exceeds $1$, this makes the cubic term contribute less. So, we finally get our equation as $$\ddot{x}=-x$$ which is also the equation of S.H.M. or simple harmonic motion.
And upon solving the integral: $$\int_{-1}^x\frac{dx}{\sqrt{{-x^2}+{3}}}=\int_2^tdt \\ \arcsin \frac{x}{\sqrt3}+\arcsin \frac{1}{\sqrt3}=t-2 \\ {x}={\sqrt3}\sin \left(t-2-\arcsin \frac{1}{\sqrt3}\right) \\ \dot{x}={\sqrt3}\cos \left(t-2-\arcsin \frac{1}{\sqrt3}\right)$$