The position vector of the particle is rt=acos⁡ωti^+asin⁡ωtj^, where a and ω are real constants of suitable dimensions. The acceleration is

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Correct answer: (A) Perpendicular to the velocity. Given that, rt=acos⁡ωti^+a sin⁡ωtj^ ∵v→=drt/dt=-aωsin⁡ωt i^+aωcos⁡ωt i^j a→=(dv→)/dt=-aω^2cos⁡ωti^-aω^2cos⁡ωtj^ a→.v→=a^2ω^3sin⁡ωtcos⁡ωt-a^2ω^3sin⁡ωtcos⁡ωt ⇒ a→.v→=0 Above result implies that acceleration is perpendicular to velocity.

The position vector of the particle is $r (t) = a \cos ω t \hat{i} + a \sin ω t \hat{j}$ , where $a$ and $ω$ are real constants of suitable dimensions. The acceleration is

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