Calculating Angular Momentum when Radius Changes for 2s

[MathewsMD]

26. A 2.0-kg block starts from rest on the positive x-axis 3.0 m from the origin and thereafter has an acceleration given by a = (4.0 m/s2)ˆi (3.0 m/s2)ˆj. At the end of 2.0 s its angular momentum about the origin is:
A. 0
B. (36kg·m2/s)kˆ
C. (+48kg·m2/s)kˆ
D. (96kg·m2/s)kˆ
E. (+96kg·m2/s)kˆ
ans: B

For this question, they essentially did:

L = | r x p |
L = (r)(m)(v)
L = (3m)(2kg)(6m/s) * I calculated v = 6 m/s to be the tangential velocity

Now, I was wondering why they didn't change the radius since it accelerates for 2 seconds. During this time period, it moves across the axis, parallel to the original rotation axis.

Δx = (1/2)(4m/s²)(2s)² and I added this length to my attempt, but it doesn't seem like the solution did. I was wondering if this is correct or if it's incorrect for future reference...

 

[haruspex]

If you want comments on the book(?) solution, please post the complete text of that.
But I would have thought the appropriate method here would be a purely vectorial one, which turns out to be very easy.

Spoiler

$\vec {\ddot x} = \vec c$
$\vec { \dot x} = \vec c t + \vec {\dot {x_0}} = \vec c t$
$\vec x = \frac 12 \vec c t^2 + \vec x_0$
$\vec L = m \vec x × \dot {\vec x} = m (\frac 12 \vec c t^2 + \vec x_0)× \vec c t = m \vec x_0 × \vec c t$