Final answer:
The particle crosses the origin at approximately 0.816 seconds and momentarily stops at 0 seconds. It passes through the origin at approximately -0.816 seconds in the negative direction and at approximately 0.816 seconds in the positive direction.
Step-by-step explanation:
(a) To find when the particle crosses the origin, we can set x(t) = 0 and solve for t. In this case, we have 4.00 - 6.00t^2 = 0. Solving for t, we get t = ±sqrt(4/6) ≈ ±0.816. Since we are dealing with time, we can ignore the negative solution. Therefore, the particle crosses the origin at t ≈ 0.816 seconds.
(b) To find where the particle momentarily stops, we need to find when the velocity is zero. The velocity function v(t) is the derivative of the position function x(t). Taking the derivative of x(t) = 4.00 - 6.00t^2, we get v(t) = -12.00t. Setting v(t) = 0, we get t = 0. Therefore, the particle momentarily stops at t = 0 seconds.
(c) To find the negative time when the particle passes through the origin, we need to find when x(t) = 0 for negative values of t. Since the position function is symmetric with respect to the origin, the particle passes through the origin at the same magnitude of time but in the negative direction, so at t ≈ -0.816 seconds.
(d) To find the positive time when the particle passes through the origin, we need to find when x(t) = 0 for positive values of t. Again, since the position function is symmetric with respect to the origin, the particle passes through the origin at the same magnitude of time but in the positive direction, so at t ≈ 0.816 seconds.