Question

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

• A. $a\omega \cos \omega t$
• B. $a\sin \omega t$
• C. $a\omega \sin \omega t$
• D. $-\frac{1}{2}(a{\omega }^2\sin \omega t)t^2$

Answer 1

Answer: (B)

Acceleration of the particle is given as $A=-a{\omega }^2\sin \omega t$ So, the velocity of the particle; $v=\int Adt$ $=\int (-a{\omega }^2\sin \omega t)dt$ $=a\omega \cos \omega t$ $\therefore$ Displacement of the particle, $x=\int vdt$ $=\int (a\omega \cos \omega t)dt$ $a\sin \omega t$