1. **Problem statement:** We have two particles A and B moving in the x-y plane. - Particle A starts at the origin (0,0) at time $t=0$ and moves at speed 10 m/s at an angle of 60° above the positive x-axis. - Particle B starts at point $(2\sqrt{3}, 9)$ at time $t=0$ and moves at speed $\frac{5}{3}$ m/s parallel to the positive x-axis. We need to find: (a) The components of velocity of particle A in x and y directions. (b) The position vector of particle A at time $t$. (c) The position vector of particle B at time $t$. (d) Show that the particles collide. 2. **Formulas and rules:** - Velocity components for a vector with magnitude $v$ and angle $\theta$ are: $$v_x = v \cos \theta, \quad v_y = v \sin \theta$$ - Position vector at time $t$ for constant velocity $\vec{v}$ and initial position $\vec{r}_0$ is: $$\vec{r}(t) = \vec{r}_0 + \vec{v} t$$ - Particles collide if their position vectors are equal at the same time $t$. 3. **Step (a): Velocity components of particle A** Given: - Speed $v = 10$ m/s - Angle $\theta = 60^\circ$ Calculate: $$v_x = 10 \cos 60^\circ = 10 \times \frac{1}{2} = 5$$ $$v_y = 10 \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$$ So, velocity components are: $$\vec{v}_A = 5 \mathbf{i} + 5\sqrt{3} \mathbf{j}$$ 4. **Step (b): Position vector of particle A at time $t$** Initial position $\vec{r}_{A0} = 0 \mathbf{i} + 0 \mathbf{j}$ Velocity $\vec{v}_A = 5 \mathbf{i} + 5\sqrt{3} \mathbf{j}$ Position vector: $$\vec{r}_A(t) = \vec{r}_{A0} + \vec{v}_A t = 5t \mathbf{i} + 5\sqrt{3} t \mathbf{j}$$ 5. **Step (c): Position vector of particle B at time $t$** Initial position: $$\vec{r}_{B0} = 2\sqrt{3} \mathbf{i} + 9 \mathbf{j}$$ Velocity: $$\vec{v}_B = \frac{5}{3} \mathbf{i} + 0 \mathbf{j}$$ Position vector: $$\vec{r}_B(t) = \vec{r}_{B0} + \vec{v}_B t = \left(2\sqrt{3} + \frac{5}{3} t\right) \mathbf{i} + 9 \mathbf{j}$$ 6. **Step (d): Show particles collide** Particles collide if $\vec{r}_A(t) = \vec{r}_B(t)$ for some $t$. Equate components: - x-components: $$5t = 2\sqrt{3} + \frac{5}{3} t$$ Rearranged: $$5t - \frac{5}{3} t = 2\sqrt{3}$$ $$\left(5 - \frac{5}{3}\right) t = 2\sqrt{3}$$ $$\frac{15}{3} - \frac{5}{3} = \frac{10}{3}$$ So: $$\frac{10}{3} t = 2\sqrt{3}$$ Solve for $t$: $$t = \frac{2\sqrt{3}}{\frac{10}{3}} = 2\sqrt{3} \times \frac{3}{10} = \frac{6\sqrt{3}}{10} = \frac{3\sqrt{3}}{5}$$ - y-components: $$5\sqrt{3} t = 9$$ Substitute $t = \frac{3\sqrt{3}}{5}$: $$5\sqrt{3} \times \frac{3\sqrt{3}}{5} = 3 \times 3 = 9$$ Since both x and y components match at the same $t$, particles collide at time $$t = \frac{3\sqrt{3}}{5} \text{ seconds}$$ and position $$\vec{r} = 5t \mathbf{i} + 5\sqrt{3} t \mathbf{j} = 3\sqrt{3} \mathbf{i} + 9 \mathbf{j}$$ **Final answers:** (a) Velocity components of A: $5 \mathbf{i}$ and $5\sqrt{3} \mathbf{j}$ (b) Position vector of A: $5t \mathbf{i} + 5\sqrt{3} t \mathbf{j}$ (c) Position vector of B: $\left(2\sqrt{3} + \frac{5}{3} t\right) \mathbf{i} + 9 \mathbf{j}$ (d) Particles collide at $t = \frac{3\sqrt{3}}{5}$ seconds at position $3\sqrt{3} \mathbf{i} + 9 \mathbf{j}$.