Step 1: Find the velocity vector $v(t)$ and acceleration vector $a(t)$. The position vector of particle P is given by $r = [(18t - 4t^3)i + t^2j]$ m. The mass of the particle is $m = 6$ kg. To find the velocity vector $v(t)$, differentiate the position vector $r(t)$ with respect to time $t$: $$v(t) = \frac{dr}{dt} = \frac{d}{dt}((18t - 4t^3)i + t^2j)$$ $$v(t) = (18 - 12t^2)i + (2t)j \text{ m/s}$$ To find the acceleration vector $a(t)$, differentiate the velocity vector $v(t)$ with respect to time $t$: $$a(t) = \frac{dv}{dt} = \frac{d}{dt}((18 - 12t^2)i + (2t)j)$$ $$a(t) = (-24t)i + (2)j \text{ m/s}^2$$ a) Find the kinetic energy of P when $t = 1$. First, find the velocity of P at $t=1$ s: $$v(1) = (18 - 12(1)^2)i + (2(1))j$$ $$v(1) = (18 - 12)i + 2j$$ $$v(1) = 6i + 2j \text{ m/s}$$ Next, calculate the magnitude of the velocity $|v(1)|$: $$|v(1)| = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} \text{ m/s}$$ The kinetic energy $KE$ is given by the formula $KE = \frac{1}{2}mv^2$. $$KE = \frac{1}{2}(6 \text{ kg})(\sqrt{40} \text{ m/s})^2$$ $$KE = \frac{1}{2}(6)(40)$$ $$KE = 3 \times 40$$ $$KE = 120 \text{ J}$$ The kinetic energy of P when $t=1$ is $\boxed{120 \text{ J}}$. b) Calculate the magnitude of the force acting on P when $t = 2$. First, find the acceleration of P at $t=2$ s: $$a(2) = (-24(2))i + (2)j$$ $$a(2) = -48i + 2j \text{ m/s}^2$$ Next, calculate the force vector $F(2)$ using Newton's second law, $F = ma$: $$F(2) = (6 \text{ kg})(-48i + 2j \text{ m/s}^2)$$ $$F(2) = -288i + 12j \text{ N}$$ Finally, calculate the magnitude of the force $|F(2)|$: $$|F(2)| = \sqrt{(-288)^2 + 12^2}$$ $$|F(2)| = \sqrt{82944 + 144}$$ $$|F(2)| = \sqrt{83088} \text{ N}$$ $$|F(2)| \approx 288.250 \text{ N}$$ The magnitude of the force acting on P when $t=2$ is $\boxed{\sqrt{83088} \text{ N} \approx 288.25 \text{ N}}$. c) Determine, to 2 decimal places, the value of $t$ for which the velocity and the acceleration vectors of P are at right angles. If two vectors are at right angles, their dot product is zero. So, $v(t) \cdot a(t) =