JEE Main & Advanced Physics Motion in a Straight Line / सरल रेखा में गति Question Bank Self Evaluation Test - Motion in a Strainght Line

  • question_answer
    The acceleration of a particle, starting from rest, varies with time according to the relation\[\text{a}=-\text{s}{{\omega }^{2}}\sin \omega \text{t}\]. The displacement of this particle at a time t will be

    A) \[\text{s sin }\omega \text{ t}\]

    B) \[\text{s }\omega \text{  cos }\omega \text{  t}\]

    C) \[\text{s }\omega \text{  sin}\omega \text{t}\]    

    D) \[-\frac{1}{2}\left( \text{s}{{\omega }^{2}}\text{ sin }\omega \text{t} \right){{\text{t}}^{2}}\]

    Correct Answer: A

    Solution :

    \[a=\frac{{{d}^{2}}x}{d{{t}^{2}}}=-s\omega \sin \omega t.\] On integrating, \[\frac{dx}{dt}=s{{\omega }^{2}}\frac{\cos \omega t}{\omega }=s\sin \omega t\] Again on integrating, we get \[x=s\omega \frac{\sin \omega t}{\omega }=s\sin \omega t\]


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