Step 1: Determine the acceleration of P at time $t$. The force acting on particle P is given by $F = [(3t^2 - 4t)i + (6t - 5)j]$ N. The particle has unit mass, so $m = 1$ kg. According to Newton's second law, $F = ma$. Since $m=1$, the acceleration $a$ is equal to the force $F$. $$a(t) = \frac{F}{m} = \frac{[(3t^2 - 4t)i + (6t - 5)j]}{1}$$ $$a(t) = (3t^2 - 4t)i + (6t - 5)j \text{ m/s}^2$$ The acceleration of P at time $t$ seconds is $\boxed{(3t^2 - 4t)i + (6t - 5)j \text{ m/s}^2}$. Step 2: Determine the velocity of P at time $t$ seconds. Velocity is the integral of acceleration with respect to time. $$v(t) = \int a(t)\,dt = \int [(3t^2 - 4t)i + (6t - 5)j]\,dt$$ $$v(t) = \left(\int (3t^2 - 4t)\,dt\right)i + \left(\int (6t - 5)\,dt\right)j$$ $$v(t) = (t^3 - 2t^2 + C_1)i + (3t^2 - 5t + C_2)j$$ We can write the constants of integration as a single vector $C = C_1 i + C_2 j$. $$v(t) = (t^3 - 2t^2)i + (3t^2 - 5t)j + C$$ We are given that at $t=3$ seconds, the velocity is $v(3) = (11i + 10j)$ m/s. Substitute $t=3$ into the velocity equation: $$v(3) = (3^3 - 2(3^2))i + (3(3^2) - 5(3))j + C$$ $$v(3) = (27 - 18)i + (27 - 15)j + C$$ $$v(3) = 9i + 12j + C$$ Equating this to the given velocity: $$11i + 10j = 9i + 12j + C$$ $$C = (11i + 10j) - (9i + 12j)$$ $$C = (11 - 9)i + (10 - 12)j$$ $$C = 2i - 2j$$ Substitute $C$ back into the velocity equation: $$v(t) = (t^3 - 2t^2)i + (3t^2 - 5t)j + (2i - 2j)$$ $$v(t) = (t^3 - 2t^2 + 2)i + (3t^2 - 5t - 2)j \text{ m/s}$$ The velocity of P at time $t$ seconds is $\boxed{(t^3 - 2t^2 + 2)i + (3t^2 - 5t - 2)j \text{ m/s}}$. Step 3: Determine the kinetic energy of P when it is moving parallel to the vector $i$. When the particle is moving parallel to the vector $i$, its $j$-component of velocity is zero. From the velocity equation in Step 2, set the $j$-component to zero: $$3t^2 - 5t - 2 = 0$$ Factor the quadratic equation: $$(3t + 1)(t - 2) = 0$$ This gives two possible values for $t$: $t = -\frac{1}{3}$ s or $t = 2$ s. Since time cannot be negative, we take $t = 2$ s. Now, find the velocity of P at $t=2$ s: $$v(2) = (2^3 - 2(2^2) + 2)i + (3(2^2) - 5(2) - 2)j$$ $$v(2) = (8 - 8 + 2)i + (12 - 10 - 2)j$$ $$v(2) = 2i + 0j$$ So, $v(2) = 2i$ m/s. The magnitude of the velocity is $|v(2)| = \sqrt{2^2 + 0^2} = 2$ m/s. The kinetic energy $KE$ is given by the formula $KE = \frac{1}{2}mv^2$. Given $m=1$ kg and $v=2$ m/s: $$KE = \frac{1}{2}(1 \text{ kg})(2 \text{ m/s})^2$$ $$KE = \frac