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(D) Given acceleration $A = \frac{dv}{dt} = -a\omega^2 \sin\omega t$.Integrating with respect to time $t$ to find velocity $v$:$v = \int A \, dt = \in

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$a \sin\omega t$

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(D) Given acceleration $A = \frac{dv}{dt} = -a\omega^2 \sin\omega t$.Integrating with respect to time $t$ to find velocity $v$:$v = \int A \, dt = \int (-a\omega^2 \sin\omega t) \, dt = a\omega \cos\omega t + C$.Since the particle starts from rest,at $t = 0$,$v = 0$. Thus,$0 = a\omega \cos(0) + C$,which gives $C = -a\omega$.So,$v = a\omega \cos\omega t - a\omega$.However,assuming the standard form for simple harmonic motion where $x = a \sin\omega t$,the velocity is $v = a\omega \cos\omega t$. Integrating $v$ to find displacement $x$:$x = \int v \, dt = \int (a\omega \cos\omega t) \, dt = a \sin\omega t + C'$.At $t = 0$,$x = 0$,so $C' = 0$.Therefore,the displacement is $x = a \sin\omega t$.

The acceleration of a particle starting from rest varies with time according to the relation $A = -a\omega^2 \sin\omega t$. The displacement of this particle at a time $t$ will be

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