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(B) The acceleration vector $\vec{a}$ can be resolved into tangential acceleration $\vec{a}_t$ and centripetal acceleration $\vec{a}_c$. Tangential ac

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accelerated circular motion

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(B) The acceleration vector $\vec{a}$ can be resolved into tangential acceleration $\vec{a}_t$ and centripetal acceleration $\vec{a}_c$. Tangential acceleration $\vec{a}_t$ is parallel to the velocity vector $\vec{v}$ (and thus parallel to momentum $\vec{p}$),while centripetal acceleration $\vec{a}_c$ is perpendicular to $\vec{v}$. We calculate the dot product of $\vec{a}$ and $\vec{p}$ to determine the nature of the speed change: $\vec{a} \cdot \vec{p} = (4\hat{i} + 3\hat{j}) \cdot (8\hat{i} - 6\hat{j}) = (4 	imes 8) + (3 	imes -6) = 32 - 18 = 14$. Since $\vec{a} \cdot \vec{p} > 0$,the angle between $\vec{a}$ and $\vec{v}$ is acute. This implies that the tangential component of acceleration is in the direction of velocity,causing the speed of the particle to increase. Therefore,the motion is accelerated circular motion.

$A$ particle is moving in a circular path. The acceleration and momentum of the particle at a certain moment are $\vec a = (4\hat i + 3\hat j)\ m/s^2$ and $\vec p = (8\hat i - 6\hat j)\ kg \cdot m/s$. The motion of the particle is

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