A particle P of mass 1 kg is subjected to a force F = [(3t2 - 4t)i + (6t - 5)j] N. If its velocity at t=3s is (11i + 10j) m/s, find its acceleration and velocity at time t.

Since the mass $m = 1$ kg, the acceleration $a$ is equal to the force $F$.

The acceleration of P at time $t$ is $\boxed{(3t^2 - 4t)i + (6t - 5)j \text{ m/s}^2}$.

Let $C = C_1 i + C_2 j$ be the constant of integration.

We are given that $v(3) = (11i + 10j)$ m/s. Substitute $t=3$ into the velocity equation:

Equate this to the given velocity at $t=3$:

The velocity of P at time $t$ is $\boxed{(t^3 - 2t^2 + 2)i + (3t^2 - 5t - 2)j \text{ m/s}}$.

Moving parallel to the vector $i$ means the $j$-component of the velocity is zero.
From part (b), the $j$-component of $v(t)$ is $(3t^2 - 5t - 2)$.
Set this to zero:

Solve the quadratic equation for $t$:

Two possible values for $t$:

Since time must be positive, we use $t=2$ s.

Substitute $t=2$ into the velocity equation from part (b):

The magnitude of the velocity is $|v(2)| = \sqrt{2^2 + 0^2} = \sqrt{4} = 2$ m/s.

The kinetic energy $KE$ is given by $KE = \frac{1}{2}mv^2$.
Given $m=1$ kg and $v=2$ m/s.

The kinetic energy of P when it is moving parallel to the vector $i$ is $\boxed{2 \text{ J}}$.

Particle Q moves with velocity $v_Q = (3i + 4j)$ m/s.

From part (c), we found $v_P(2) = 2i$ m/s.

The velocity of P relative to Q is $v_{P/Q} = v_P - v_Q$.

The velocity of P relative to Q at time $t=2$ is $\boxed{-i - 4j \text{ m/s}}$.