What is the motion of the block when released from a compressed spring?

[allyfranken]

Homework Statement

The 15.8-in. spring is compressed to a 7.1-in. length, where it is released from rest and accelerates the sliding block A. The acceleration has an initial value of 160 ft/sec2 and then decreases linearly with the x-movement of the block, reaching zero when the spring regains its original 15.8-in. length. Calculate the time t for the block to go (a) 4.35 in. and (b) 8.7 in.

Homework Equations

a = 160 - kx

vdV = adX
V = dx/dt

The Attempt at a Solution

k = 160/((15.8-7.1)/12)

a = 160 - 220X

Then i integrated vDV = 160 - 220x
and got: v^2 = 320x - 220x^2
solved for V = sqrt(320x - 220x^2)

now I know that V = dx/dt and to solve for dt and integrate for T. However I get stuck at the integral of dx/sqrt(320x - 220x^2) so I assume I am doing something wrong in the process of getting there.

 

[haruspex]

If the block were attached to the end of the spring and the system allowed to continue to move, what kind of motion would you see? What does that suggest for the form of x as a function of t?