A particle moves in the xy-plane with a constant acceleration given by vec{a} = 3hat{i} + 4hat{j}text{ m/s}^2. At t = 0, the particle starts from the…

Using $\vec{v} = \vec{v}_0 + \vec{t}$, we have $\vec{v} = 2\hat{i} + (3\hat{i} + 4\hat{j})(2) = 2\hat{i} + 6\hat{i} + 8\hat{j} = 8\hat{i} + 8\hat{j}$. The magnitude is $\sqrt{8^2 + 8^2} = \sqrt{128} \approx 11.3$ wait, recalculate: $8\hat{i} + 8\hat{j}$ gives magnitude $\sqrt{64+64}=\sqrt{128}$. Let's recheck initial data: $v_x = 2 + 3(2) = 8$, $v_y = 0 + 4(2) = 8$. Magnitude is $\sqrt{64+64} = 8\sqrt{2}\text{ m/s}$. Let's adjust options or numbers. Let $\vec{a} = 3\hat{i} + 4\hat{j}$ and $\vec{v}_0 = -2\hat{i}$? Let's fix with nice numbers: $\vec{v}_0 = 2\hat{i}$ and $\vec{a} = 3\hat{i} + 2\hat{j}$ and $t=2$. $\vec{v} = 2\hat{i} + (3\hat{i}+2\hat{j})2 = 8\hat{i} + 4\hat{j}$. Magnitude $\sqrt{64+16}=\sqrt{80}$. Let's rewrite question so answer is 10: $\vec{v}_0 = \hat{i}$, $\vec{a} = 2\hat{i} + 3\hat{j}$, $t=2$. $\vec{v} = \hat{i} + (4\hat{i} + 6\hat{j}) = 5\hat{i} + 6\hat{j}$ (no). Let's use $\vec{v}_0 = 2\hat{i} - 2\hat{j}$, $\vec{a} = 4\hat{i} + 3\hat{j}$, $t=2$. $\vec{v} = 2\hat{i} - 2\hat{j} + 8\hat{i} + 6\hat{j} = 10\hat{i} + 4\hat{j}$. Magnitude $=\sqrt{100+16}=\sqrt{116}$. Let's stick to standard clean values: If $\vec{v}_0 = 2\hat{i} + 3\hat{j}$ and $\vec{a} = 1\hat{i} + 2\hat{j}$, at $t=2$, $\vec{v} = 4\hat{i} + 7\hat{j}$. Let's make a simpler one: $\vec{v} = \vec{v}_0 + \vec{a}t$. Let $\vec{v}_0 = 0$, $\vec{a} = 3\hat{i} + 4\hat{j}$, $t=2$, $\vec{v} = 6\hat{i} + 8\hat{j}$, magnitude $\sqrt{36+64} = 10\text{ m/s}$.