The equation for the acceleration of a particle is given by a = 2 + 2t.

Velocity $v(t)$ is the integral of acceleration $a(t)$ with respect to time $t$:

$$v(t) = \int a(t)\,dt$$

Given $a(t) = 2 + 2t$:

$$v(t) = \int (2 + 2t)\,dt$$ $$v(t) = \int 2\,dt + \int 2t\,dt$$ $$v(t) = 2t + 2\left(\frac{t^2}{2}\right) + C$$ $$v(t) = 2t + t^2 + C$$

Here, $C$ is the constant of integration, representing the initial velocity $v(0)$. Since no initial velocity is given, we assume the particle starts from rest, meaning $v(0) = 0$.
Substituting $t=0$ and $v(0)=0$:

$$0 = 2(0) + (0)^2 + C$$ $$C = 0$$

So, the velocity function is:

$$v(t) = 2t + t^2$$

Substitute $t = 4$ into the velocity function:

$$v(4) = 2(4) + (4)^2$$ $$v(4) = 8 + 16$$ $$v(4) = 24\text{m/s}$$

Comparing this result with the given options:
A. $8\text{m/s}$
B. $32\text{m/s}$
C. $24\text{m/s}$
D. $42\text{m/s}$

The calculated instantaneous velocity matches option C.

The final answer is $\boxed{\text{C. *24 m/s}}$.