non constant acceleration problem 
The acceleration of an arrow from a bow falls from $6000m/s^2$ to zero when it leaves the bow after travelling a distance $x=0.75m$. Assuming that this acceleration can be expressed by the linear equation $a=6000[1-(4x/3)] m/s^2$ determine the speed of the arrow when it leaves the bow.
How long does it take for the arrow to leave the bow?

to answer the first part of the question
$v dv = a dx$
$\int_U^V{v}$ dv = $\int_0^x{6000(1-4x/3)}$ dx
$\frac{(V^2 - U^2)}{2}$ = $2000(3x-2x^2)$
$V^2 = 4000(3x-2x^2) + U^2$
$V = \sqrt{4000(3x-2x^2) +U^2}$
substitute $x= 0.75$ and $U=0$ giving,
$$v=67.1 m/s$$
For the second part of the question I have got this far
$v=\frac{dx}{dt}$
$dt = \frac{1}{v} dx$
$\int_0^T{dt} =\int_0^x \frac{1}{v} dx$
$T = \int_0^x \frac{1}{\sqrt{4000(3x-2x^2) +U^2}} dx$
this integral is nasty, giving a complex solution but works out when I solve it giving the correct answer of $$t=0.0176s$$
Can anyone see a more efficient method for solving this problem? Is there a substitution I could use to make the integral easier? or am I missing an aspect of this question.
 A: This is the setup described in the equation:

The acceleration is defined in terms os the displacement of the bow $x$ by:
$$ a = 6000 \left(1 - \tfrac{4}{3}x\right) \tag{1} $$
So initially $x=0$ and when we substitute this into equation (1) we get $a = 6000 \text{ms}^{-2}$. When the arrow leaves the bow so $x=\tfrac{3}{4}$ and we get $a=0$. So far so good.
But suppose we choose a different definition for the variable $x$ as shown below:

So now $x$ starts at $\tfrac{3}{4}$m and when the arrow leaves the bow $x=0$. If we define $x$ this way then the equation for the acceleration becomes:
$$ a = -8000x \tag{2} $$
let's just check this: at the start $x = \tfrac{3}{4}$m and putting this into equation (2) gives $a = 6000 \text{ms}^{-2}$. When the arrow leaves the bow $x=0$ and equation (2) gives $a=0$.
So equation (2) gives us the acceleration with our redefined meaning for $x$. But equation (2) is just the equation of motion for a simple harmonic oscillator:
$$ \frac{d^2x}{dt^2} = -kx $$
So the motion of the arrow is going to be given by an equation:
$$ x = \tfrac{3}{4}\cos\left(2\pi \frac{t}{\tau}\right) $$
where you can calculate the period $\tau$ by solving equation (2).
