Position is the integral of velocity. $$x(t) = \int v_x(t) dt = \int(4t^3 - 2t) dt = t^4 - t^2 + C$$. Use the condition $$x(1)=1$$ to find $$C$$. $$1 = (1)^4 - (1)^2 + C$$, which gives $$1 = 0 + C$$, so $$C=1$$. The position function is $$x(t) = t^4 - t^2 + 1$$. At $$t=2$$, the position is $$x(2) = (2)^4 - (2)^2 + 1 = 16 - 4 + 1 = 13$$.

The object is momentarily at rest when its velocity is zero. We find the velocity function by differentiating the position function: $$v(t) = \frac{dx}{dt} = 6t^2 - 30t + 36$$. Set $$v(t) = 0$$ and solve for $$t$$: $$6(t^2 - 5t + 6) = 0$$, which factors to $$6(t-2)(t-3) = 0$$. The solutions are $$t=2 \text{ s}$$ and $$t=3 \text{ s}$$.

To find position from acceleration, we must integrate twice. First, find velocity: $$v_x(t) = \int a_x(t) dt = \int C t^{1/2} dt = C \frac{t^{3/2}}{3/2} + v_0$$. Since the particle starts from rest, $$v_0=0$$, so $$v_x(t) = \frac{2C}{3}t^{3/2}$$. Next, find position: $$x(t) = \int v_x(t) dt = \int \frac{2C}{3}t^{3/2} dt = \frac{2C}{3} \frac{t^{5/2}}{5/2} + x_0$$. Since it starts at the origin, $$x_0=0$$. Thus, $$x(t) = \frac{4C}{15}t^{5/2}$$.

Displacement is the definite integral of velocity over the time interval. $$\Delta x = \int_{1}^{3} v(t) dt = \int_{1}^{3} (3t^2 - 6t) dt$$. Evaluating the integral gives $$[t^3 - 3t^2]_{1}^{3} = ((3)^3 - 3(3)^2) - ((1)^3 - 3(1)^2) = (27 - 27) - (1 - 3) = 0 - (-2) = 2 \text{ m}$$.

Instantaneous velocity is the time derivative of the position function, $$v(t) = \frac{dx}{dt}$$. Differentiating $$x(t)$$ with respect to time gives $$v(t) = 3At^2 - 2Bt$$. Substituting the given values and $$t=2.0 \text{ s}$$ yields $$v(2.0) = 3(2.0)(2.0)^2 - 2(3.0)(2.0) = 24 - 12 = 12 \text{ m/s}$$.

First, find the time when the particle momentarily stops by setting $$v(t) = 0$$. This gives $$12t - 3t^2 = 3t(4 - t) = 0$$, so $$t=0 \text{ s}$$ and $$t=4 \text{ s}$$. Acceleration is the derivative of velocity: $$a(t) = \frac{dv}{dt} = 12 - 6t$$. At $$t=4 \text{ s}$$, the acceleration is $$a(4) = 12 - 6(4) = 12 - 24 = -12 \text{ m/s}^2$$.

Velocity is the integral of acceleration with respect to time. $$v(t) = \int a(t) dt = \int(6t - 4) dt = 3t^2 - 4t + C$$. We use the initial condition $$v(0) = 5 \text{ m/s}$$ to find the constant of integration $$C$$. $$v(0) = 3(0)^2 - 4(0) + C = 5$$, so $$C=5$$. The velocity function is $$v(t) = 3t^2 - 4t + 5$$. At $$t=3 \text{ s}$$, $$v(3) = 3(3)^2 - 4(3) + 5 = 27 - 12 + 5 = 20 \text{ m/s}$$.

At the peak of its trajectory, a vertically thrown ball momentarily stops moving upwards before it begins to fall back down, so its instantaneous velocity is zero. However, the acceleration due to gravity is still acting on it, so its acceleration is approximately $$-9.8 \text{ m/s}^2$$. Therefore, it has zero velocity and non-zero acceleration.

A particle is slowing down when its velocity and acceleration have opposite signs. First, find $$v(t) = \frac{dx}{dt} = 3t^2 - 12t = 3t(t-4)$$. Then find $$a(t) = \frac{dv}{dt} = 6t - 12 = 6(t-2)$$. We analyze the signs: For $$t \in(2, 4)$$, $$v(t)$$ is negative (e.g., at $$t=3$$, $$v=27-36=-9$$) and $$a(t)$$ is positive (e.g., at $$t=3$$, $$a=18-12=6$$). Since they have opposite signs, the particle is slowing down in this interval.

Average velocity is defined as displacement divided by the time interval: $$v_{avg} = \frac{\Delta x}{\Delta t} = \frac{x(3) - x(1)}{3-1}$$. First, calculate the positions: $$x(3) = 2(3)^3 - 3(3)^2 + 4 = 54 - 27 + 4 = 31 \text{ m}$$. $$x(1) = 2(1)^3 - 3(1)^2 + 4 = 2 - 3 + 4 = 3 \text{ m}$$. Then, $$v_{avg} = \frac{31 - 3}{2} = \frac{28}{2} = 14 \text{ m/s}$$.