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(D) Given the tangential acceleration $a_t = K^2rt^2$.Since $a_t = \frac{dv}{dt}$,we have $dv = K^2rt^2 dt$.Integrating both sides,$\int dv = \int K^2

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Power delivered by centripetal force will always be zero.

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(D) Given the tangential acceleration $a_t = K^2rt^2$.Since $a_t = \frac{dv}{dt}$,we have $dv = K^2rt^2 dt$.Integrating both sides,$\int dv = \int K^2rt^2 dt$,which gives $v = \frac{K^2rt^3}{3}$.Since the velocity $v$ changes with time,the centripetal acceleration $a_c = \frac{v^2}{r}$ is not constant.The power delivered by a force $\vec{F}$ is given by $P = \vec{F} \cdot \vec{v}$.The centripetal force $\vec{F}_c$ is always directed towards the center of the circular path,while the velocity vector $\vec{v}$ is always tangential to the path.Therefore,the angle between $\vec{F}_c$ and $\vec{v}$ is always $90^\circ$.Thus,the power delivered by the centripetal force is $P_c = \vec{F}_c \cdot \vec{v} = F_c v \cos(90^\circ) = 0$.

$A$ particle of mass $m$ is moving in a circular path of constant radius $r$ such that its tangential acceleration varies with time as $a_t = K^2rt^2$ ($K$ is a constant). Select the correct statement.

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