First, we recognize that the velocity of a particle at a moment in time can be found by integrating the acceleration function with respect to time. This means we need to find the integral of the acceleration function aₓ(t) = -2.01 m/s² + 2.98 m/s²*t over the interval [0, t].
The definite integral of aₓ(t) = -2.01 m/s² + 2.98 m/s²*t from t=0 to any particular t gives us the velocity at that time. Evaluating this integral at t=0 gives us the initial velocity, v₀ₓ. Performing this computation, we find that the initial velocity v₀ₓ equals to 0 m/s².
However, it's given that we want the particle to have the same x-coordinate at t = 3.91s as it had at t=0. This means that the total displacement of the particle over this time must be zero. Recall that displacement equals to the area under the velocity-time graph, so this is equivalent to requiring the integral from 0 to 3.91s of the velocity function be zero.
In other words, the average velocity over this time period needs to be zero. To ensure this, we adjust the initial velocity, v₀ₓ to fulfill this condition. We subtract the integral from 0 to t (3.91s) of the velocity function, divided by 3.91s. Performing this adjustment using t = 3.91, we find that the adjusted initial velocity is -3.777741 m/s.
So, the initial velocity is 0 m/s, and if we want the particle to return to its initial position at t = 3.91, we need to adjust this by -3.777741 m/s.