The average velocity of a particle from t=0 to t=3 sec can be calculated by finding the change in position divided by the change in time. In this case, we need to calculate the average velocity using the given position equation y=2t^3 - 3t + 4.

To find the change in position, we substitute the values of t=3 and t=0 into the position equation and calculate the corresponding positions at those times.

Substituting t=3:
y(3) = 2(3)^3 - 3(3) + 4
= 2(27) - 9 + 4
= 54 - 9 + 4
= 49

Substituting t=0:
y(0) = 2(0)^3 - 3(0) + 4
= 0 - 0 + 4
= 4

The change in position is given by the difference between the final and initial positions:
Change in position = y(3) - y(0)
= 49 - 4
= 45

The change in time is simply t=3 - t=0 = 3 - 0 = 3.

Now, we can calculate the average velocity by dividing the change in position by the change in time:
Average velocity = (Change in position) / (Change in time)
= 45 / 3
= 15

Therefore, the average velocity of the particle from t=0 to t=3 sec is 15 m/s.

However, the given answer is 9 m/s. To investigate this discrepancy, let's recheck the position equation.

The given position equation is y=2t^3 - 3t + 4. Taking its derivative with respect to time, we can find the instantaneous velocity at any given time.

dy/dt = 6t^2 - 3

Substituting t=3 into the velocity equation, we can find the instantaneous velocity at t=3.

v(3) = 6(3)^2 - 3
= 6(9) - 3
= 54 - 3
= 51

So, the instantaneous velocity at t=3 sec is 51 m/s. This means that the average velocity cannot be 9 m/s, as it is lower than the instantaneous velocity at t=3 sec.

Therefore, there seems to be an error in either the given answer or the provided position equation.