Given the acceleration function a = 2t2 + t − 4, the velocity v is found by integrating a with respect to t:

Given the acceleration function $a = 2t^2 + t - 4$, the velocity $v$ is found by integrating $a$ with respect to $t$:

$$v = \int a\,dt$$ $$v = \int (2t^2 + t - 4)\,dt$$

Integrate each term of the acceleration function:

$$v = \frac{2t^{2+1}}{2+1} + \frac{t^{1+1}}{1+1} - 4t + C$$ $$v = \frac{2}{3}t^3 + \frac{1}{2}t^2 - 4t + C$$

Here, $C$ is the constant of integration.

The problem states that $v=0$ when $t=0$. Substitute these values into the velocity equation:

$$0 = \frac{2}{3}(0)^3 + \frac{1}{2}(0)^2 - 4(0) + C$$ $$0 = 0 + 0 - 0 + C$$ $$C = 0$$

Substitute the value of $C$ back into the velocity equation:

$$v = \frac{2}{3}t^3 + \frac{1}{2}t^2 - 4t$$

The velocity $v$ is expressed in $\text{m/s}$.

The velocity of the particle is:

$$v = \frac{2}{3}t^3 + \frac{1}{2}t^2 - 4t \text{ m/s}$$