1. **State the problem:** A particle of mass 2.5 kg is acted upon by its weight and two external forces \(\mathbf{F}_1 = 3\mathbf{i} - 2\mathbf{j}\) N and \(\mathbf{F}_2 = -\mathbf{i} + 18\mathbf{j}\) N. We need to find the acceleration vector of the particle. 2. **Relevant formula:** Newton's second law states that the net force \(\mathbf{F}_{net}\) on an object is equal to its mass \(m\) times its acceleration \(\mathbf{a}\): $$\mathbf{F}_{net} = m \mathbf{a}$$ Therefore, $$\mathbf{a} = \frac{\mathbf{F}_{net}}{m}$$ 3. **Calculate the weight force:** Weight \(\mathbf{W}\) acts downward (negative \(\mathbf{j}\) direction) and is given by: $$\mathbf{W} = m \mathbf{g} = 2.5 \times (0\mathbf{i} - 9.8\mathbf{j}) = 0\mathbf{i} - 24.5\mathbf{j} \text{ N}$$ 4. **Calculate the net force:** Sum all forces: $$\mathbf{F}_{net} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{W} = (3\mathbf{i} - 2\mathbf{j}) + (-\mathbf{i} + 18\mathbf{j}) + (0\mathbf{i} - 24.5\mathbf{j})$$ Simplify components: $$\mathbf{F}_{net} = (3 - 1 + 0)\mathbf{i} + (-2 + 18 - 24.5)\mathbf{j} = 2\mathbf{i} - 8.5\mathbf{j} \text{ N}$$ 5. **Calculate acceleration:** $$\mathbf{a} = \frac{\mathbf{F}_{net}}{m} = \frac{2\mathbf{i} - 8.5\mathbf{j}}{2.5} = 0.8\mathbf{i} - 3.4\mathbf{j} \text{ m/s}^2$$ **Final answer:** The acceleration of the particle in vector form is: $$\boxed{\mathbf{a} = 0.8\mathbf{i} - 3.4\mathbf{j} \text{ m/s}^2}$$