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(D) Given force $\overrightarrow{F} = (6t \hat{i} + 6t^2 \hat{j}) \ N$ and mass $m = 2 \ kg$.Using Newton's second law,$\overrightarrow{a} = \frac{\ov

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$(9t^3 + 6t^5) \ W$

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(D) Given force $\overrightarrow{F} = (6t \hat{i} + 6t^2 \hat{j}) \ N$ and mass $m = 2 \ kg$.Using Newton's second law,$\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{6t \hat{i} + 6t^2 \hat{j}}{2} = (3t \hat{i} + 3t^2 \hat{j}) \ m/s^2$.To find velocity $\overrightarrow{v}$,we integrate acceleration with respect to time: $\overrightarrow{v} = \int \overrightarrow{a} \ dt = \int (3t \hat{i} + 3t^2 \hat{j}) \ dt = (\frac{3t^2}{2} \hat{i} + t^3 \hat{j}) \ m/s$.Power $P$ is the dot product of force and velocity: $P = \overrightarrow{F} \cdot \overrightarrow{v} = (6t \hat{i} + 6t^2 \hat{j}) \cdot (\frac{3t^2}{2} \hat{i} + t^3 \hat{j})$.$P = (6t \cdot \frac{3t^2}{2}) + (6t^2 \cdot t^3) = 9t^3 + 6t^5 \ W$.

$A$ body of mass $2 \ kg$ begins to move under the action of a time-dependent force given by $\overrightarrow{F} = (6t \hat{i} + 6t^2 \hat{j}) \ N$. The power developed by the force at time $t$ is given by:

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