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(C) Given,the position vector is $r = a \cos \omega t \hat{i} + b \sin \omega t \hat{j}$.To find the velocity $v$,we differentiate $r$ with respect to

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$-r$

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(C) Given,the position vector is $r = a \cos \omega t \hat{i} + b \sin \omega t \hat{j}$.To find the velocity $v$,we differentiate $r$ with respect to time $t$:$v = \frac{dr}{dt} = \frac{d}{dt}(a \cos \omega t \hat{i} + b \sin \omega t \hat{j}) = -a \omega \sin \omega t \hat{i} + b \omega \cos \omega t \hat{j}$.To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:$a = \frac{dv}{dt} = \frac{d}{dt}(-a \omega \sin \omega t \hat{i} + b \omega \cos \omega t \hat{j}) = -a \omega^2 \cos \omega t \hat{i} - b \omega^2 \sin \omega t \hat{j}$.Factoring out $-\omega^2$,we get:$a = -\omega^2 (a \cos \omega t \hat{i} + b \sin \omega t \hat{j})$.Since the term in the parenthesis is the original position vector $r$,we have:$a = -\omega^2 r$.This indicates that the acceleration vector is directed along $-r$.

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