A particle oscillates on the x-axis. It's position $\displaystyle x(t)$ (in meters) at the time $\displaystyle t$ (in seconds) is related to its speed $\displaystyle v(t)$ by the equation:

$\displaystyle x^2+x \cdot v=v^2=13$

Give the acceleration of this particle when its position is $\displaystyle x = -3 meters$, knowing that at that moment, the speed is positive.

Hint: This is a related rates problem in which you need to use the implicit derivative. Use the fact that $\displaystyle a(t) = \frac{dv}{dt}$ and $\displaystyle v(t) = \frac{dx}{dt}$