1. **Problem statement:** A particle starts from rest at point O and accelerates with vector acceleration $\mathbf{a} = 5\mathbf{i} - 11\mathbf{j} + 2\mathbf{k}$ m/s$^2$ for 4 seconds. We need to find: a) The speed of the particle after 4 seconds in the form $a\sqrt{6}$ m/s. b) The distance travelled by the particle in the first 4 seconds. 2. **Relevant formulas and rules:** - Velocity after time $t$ when starting from rest: $\mathbf{v} = \mathbf{a}t$ - Speed is the magnitude of velocity: $v = |\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$ - Distance travelled under constant acceleration from rest: $s = \frac{1}{2} a t^2$ where $a$ is the magnitude of acceleration. 3. **Calculate velocity vector after 4 seconds:** $$\mathbf{v} = (5\mathbf{i} - 11\mathbf{j} + 2\mathbf{k}) \times 4 = 20\mathbf{i} - 44\mathbf{j} + 8\mathbf{k}$$ 4. **Calculate speed (magnitude of velocity):** $$v = \sqrt{20^2 + (-44)^2 + 8^2} = \sqrt{400 + 1936 + 64} = \sqrt{2400}$$ 5. **Simplify $\sqrt{2400}$:** $$\sqrt{2400} = \sqrt{400 \times 6} = \sqrt{400} \times \sqrt{6} = 20\sqrt{6}$$ So, $a = 20$ and speed after 4 seconds is $20\sqrt{6}$ m/s. 6. **Calculate distance travelled:** - First find magnitude of acceleration: $$a = \sqrt{5^2 + (-11)^2 + 2^2} = \sqrt{25 + 121 + 4} = \sqrt{150} = 5\sqrt{6}$$ - Use formula for distance under constant acceleration from rest: $$s = \frac{1}{2} a t^2 = \frac{1}{2} \times 5\sqrt{6} \times 4^2 = \frac{1}{2} \times 5\sqrt{6} \times 16 = 40\sqrt{6}$$ **Final answers:** - a) Speed after 4 seconds: $20\sqrt{6}$ m/s - b) Distance travelled in 4 seconds: $40\sqrt{6}$ m