A particle is moving in the xy-plane such tha | vedclass

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(B) The position vector of the particle is given by $\vec{r} = x\hat{i} + y\hat{j} = (4t + t^2)\hat{i} + (2t + \frac{t^2}{2})\hat{j}$.Velocity is the

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$\vec{v} = (4 + 2t) \hat{i} + (2 + t) \hat{j}$

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(B) The position vector of the particle is given by $\vec{r} = x\hat{i} + y\hat{j} = (4t + t^2)\hat{i} + (2t + \frac{t^2}{2})\hat{j}$.Velocity is the time derivative of the position vector: $\vec{v} = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}$.Differentiating $x$ with respect to $t$: $v_x = \frac{d}{dt}(4t + t^2) = 4 + 2t$.Differentiating $y$ with respect to $t$: $v_y = \frac{d}{dt}(2t + \frac{t^2}{2}) = 2 + t$.Therefore, the velocity vector is $\vec{v} = (4 + 2t)\hat{i} + (2 + t)\hat{j} 	ext{ m/s}$.

$A$ particle is moving in the $xy$-plane such that its position coordinates are given by $x = (4t + t^2) \text{ m}$ and $y = (2t + \frac{t^2}{2}) \text{ m}$, where $t$ is in seconds. What is the velocity of the particle?

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