The work done $W$ by the force $\vec{F_a}$ is given by $W = \int F_a dx$. Using Newton's second law, $F_a = ma$, the work can be expressed as:

$$W = \int (ma) dx = m \int a dx$$

The integral $\int a dx$ represents the area under the acceleration-position ($a-x$) graph. The work-energy theorem states that the net work done on an object equals its change in kinetic energy, $W = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$. Since the particle starts from rest, $v_i = 0$, so $W = \frac{1}{2}mv_f^2$. This allows us to find the final speed $v_f$:

$$v_f = \sqrt{\frac{2W}{m}} = \sqrt{2 \int a dx}$$

We calculate the area under the $a-x$ graph for each position. Let $A(x) = \int_0^x a dx'$. Area from $x=0$ to $x=4.0$ m: This is a trapezoid, broken into a triangle (0 to 1 m) and a rectangle (1 to 4 m).

$$A(4) = \left(\frac{1}{2} \cdot 1.0 \cdot a_s\right) + \left(3.0 \cdot a_s\right) = 0.5 a_s + 3.0 a_s = 3.5 a_s$$ $$A(4) = 3.5 \cdot (6.0 \text{ m/s}^2) = 21.0 \text{ m}^2\text{/s}^2$$

Area from $x=0$ to $x=7.0$ m: We add the area from 4 m to 7 m to $A(4)$. This consists of a positive triangle (4 to 5 m), a negative triangle (5 to 6 m), and a negative rectangle (6 to 7 m).

$$A(7) = A(4) + \left(\frac{1}{2} \cdot 1.0 \cdot a_s\right) + \left(\frac{1}{2} \cdot 1.0 \cdot (-a_s)\right) + \left(1.0 \cdot (-a_s)\right) = A(4) - a_s$$ $$A(7) = 21.0 - 6.0 = 15.0 \text{ m}^2\text{/s}^2$$

Area from $x=0$ to $x=9.0$ m: We add the area from 7 m to 9 m to $A(7)$. This consists of a negative rectangle (7 to 8 m) and a negative triangle (8 to 9 m).

$$A(9) = A(7) + \left(1.0 \cdot (-a_s)\right) + \left(\frac{1}{2} \cdot 1.0 \cdot (-a_s)\right) = A(7) - 1.5 a_s$$ $$A(9) = 15.0 - 1.5(6.0) = 15.0 - 9.0 = 6.0 \text{ m}^2\text{/s}^2$$

[Q1-Q3] The work done is $W(x) = m \cdot A(x)$:

$W(4) = (2.00 \text{ kg})(21.0 \text{ m}^2\text{/s}^2) = 42.0 \text{ J}$ $W(7) = (2.00 \text{ kg})(15.0 \text{ m}^2\text{/s}^2) = 30.0 \text{ J}$ $W(9) = (2.00 \text{ kg})(6.0 \text{ m}^2\text{/s}^2) = 12.0 \text{ J}$

[Q4-Q6] The speed is $v(x) = \sqrt{2 \cdot A(x)}$:

$v(4) = \sqrt{2 \cdot 21.0} = \sqrt{42.0} \approx 6.48 \text{ m/s}$ $v(7) = \sqrt{2 \cdot 15.0} = \sqrt{30.0} \approx 5.48 \text{ m/s}$ $v(9) = \sqrt{2 \cdot 6.0} = \sqrt{12.0} \approx 3.46 \text{ m/s}$

Since the cumulative area is always positive, the kinetic energy is always positive and the particle never comes to rest. Therefore, its direction of travel is always in the positive x-direction.