Originally Posted by

**Gusbob** **Question:**

The acceleration of a particle is given by a = 3x+1 (where a is acceleration, x is displacement from origin). If the particle is initially at the origin and moving at a velocity of $\displaystyle 15ms^{-1}$, find its velocity after 3s.

**What I've done:**

$\displaystyle a = v\frac{dv}{dx} = 3x -1$

$\displaystyle \frac{1}{2} v^2 = \frac{3}{2}x^2 + x + c $

when t=0, x=0, v = 15

$\displaystyle v^2 = 3x^2 + 2x + 225$

I'm not sure where to go on from here. I know if I find t in terms of x, I can substitute it into that equation. I've tried $\displaystyle \frac{dx}{dt} = \sqrt{3x^2 + 2x + 225} \Rightarrow \frac{dt}{dx} = \frac{1}{\sqrt{3x^2 + 2x + 225}}$ but it doesn't work well.