Here is the solution to question 1(i): 1(i) A force $\mathbf{F} = (18\mathbf{i} + 18\mathbf{j} - 36\mathbf{k})$ N acts at time $t$ seconds on a particle of mass 3kg. Initially, the particle is at the origin moving with velocity $(2\mathbf{i} - \mathbf{j} - 3\mathbf{k})$ ms$^{-1}$. a) Find the acceleration of the particle at time $t$. Step 1: Use Newton's second law, $\mathbf{F} = m\mathbf{a}$. Given force $\mathbf{F} = (18\mathbf{i} + 18\mathbf{j} - 36\mathbf{k})$ N and mass $m = 3$ kg. $$ \mathbf{a} = \frac{\mathbf{F}}{m} $$ Step 2: Substitute the given values. $$ \mathbf{a} = \frac{1}{3}(18\mathbf{i} + 18\mathbf{j} - 36\mathbf{k}) $$ Step 3: Simplify to find the acceleration. $$ \mathbf{a} = (6\mathbf{i} + 6\mathbf{j} - 12\mathbf{k})\, \text{ms}^{-2} $$ The acceleration of the particle is $\boxed{(6\mathbf{i} + 6\mathbf{j} - 12\mathbf{k})\, \text{ms}^{-2}}$. b) Find the velocity of the particle at time $t$. Step 1: Use the kinematic equation for velocity, $\mathbf{v} = \mathbf{u} + \mathbf{a}t$. Given initial velocity $\mathbf{u} = (2\mathbf{i} - \mathbf{j} - 3\mathbf{k})$ ms$^{-1}$ and acceleration $\mathbf{a} = (6\mathbf{i} + 6\mathbf{j} - 12\mathbf{k})$ ms$^{-2}$ (from part a). Step 2: Substitute the values into the equation. $$ \mathbf{v} = (2\mathbf{i} - \mathbf{j} - 3\mathbf{k}) + (6\mathbf{i} + 6\mathbf{j} - 12\mathbf{k})t $$ Step 3: Group the components. $$ \mathbf{v} = (2+6t)\mathbf{i} + (-1+6t)\mathbf{j} + (-3-12t)\mathbf{k}\, \text{ms}^{-1} $$ The velocity of the particle at time $t$ is $\boxed{(2+6t)\mathbf{i} + (-1+6t)\mathbf{j} + (-3-12t)\mathbf{k}\, \text{ms}^{-1}}$. c) When $t=2$, find the kinetic energy of the particle. Step 1: Find the velocity vector at $t=2$ s using the expression from part b. $$ \mathbf{v}(2) = (2+6(2))\mathbf{i} + (-1+6(2))\mathbf{j} + (-3-12(2))\mathbf{k} $$ $$ \mathbf{v}(2) = (2+12)\mathbf{i} + (-1+12)\mathbf{j} + (-3-24)\mathbf{k} $$ $$ \mathbf{v}(2) = 14\mathbf{i} + 11\mathbf{j} - 27\mathbf{k}\, \text{ms}^{-1} $$ Step 2: Calculate the magnitude of the velocity vector, $|\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$. $$ |\mathbf{v}(2)| = \sqrt{14^2 + 11^2 + (-27)^2} $$ $$ |\mathbf{v}(2)| = \sqrt{196 + 121 + 729} $$ $$ |\mathbf{v}(2)| = \sqrt{1046}\, \text{m/s} $$ Step 3: Calculate the kinetic energy using the formula $KE = \frac{1}{2}mv^2$. Given mass $m = 3$ kg. $$ KE = \frac{1}{2}(3)(\sqrt{1046})^2 $$ $$ KE = \frac{3}{2}(1046) $$ $$ KE = 3 \times 523 $$ $$ KE = 1569\, \text{J} $$ The kinetic energy of the particle at $t=2$ s is $\boxed{1569\, \text{J}}$. d) A particle starts from rest and moves with an acceleration of $(2v^2+1)$ ms$^{-2}$ where $v$ is the speed of the particle. Find the distance taken by the particle to attain a speed of 3m/s. Step 1: Relate acceleration to velocity and distance using $a = v\frac{dv}{dx}$. Given acceleration $a = 2v^2+1$. $$ v\frac{dv}{dx} = 2v^2+1 $$ Step 2: Separate variables and integrate. $$ \frac{v}{2v^2+1}\,dv = dx $$ Integrate from initial conditions ($v=0$ at $x=0$) to final conditions ($v=3$ at $x$). $$ \int_{0}^{3} \frac{v}{2v^2+1}\,dv = \int_{0}^{x} dx $$ Step 3: Evaluate the integral. For the left side, let $u = 2v^2+1$, so $du = 4v\,dv \implies v\,dv = \frac{1}{4}\,du$. When $v=0$, $u = 2(0)^2+1 = 1$. When $v=3$, $u = 2(3)^2+1 = 18+1 = 19$. $$ \int_{1}^{19} \frac{1}{4u}\,du = [x]_{0}^{x} $$ $$ \frac{1}{4}[\ln|u|]_{1}^{19} = x $$ $$ \frac{