A particle moves along an arc of a circle of radius $R = 1.00 ext{ m}$ according to the law $l = a

Question

A particle moves along an arc of a circle of radius $R = 1.00 	ext{ m}$ according to the law $l = a \sin \omega t$, where $l$ is the displacement from the initial position measured along the arc, $a = 0.80 	ext{ m}$, and $\omega = 2.00 	ext{ rad/s}$. The total acceleration of the particle is a vector sum of its tangential and normal accelerations.

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See detailed solution for explanation.