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Given, r=\cos \omega ti+\sin \omega tj\,

i).Velocity=v={\frac  {dr}{dt}}=-\omega \sin \omega ti+\omega \cos \omega tj\,

Therefore,r\cdot {\frac  {dr}{dt}}=(\cos \omega ti+\sin \omega tj)\cdot (-\omega \sin \omega ti+\omega \cos \omega tj)\,

or r\cdot {\frac  {dr}{dt}}=-\omega \cos \omega t+\omega \sin \omega t\cos \omega t=0\,

Showing that the velocity of the particle v is perpendicular to r.

ii).Acceleration a= {\frac  {d^{2}r}{dt^{2}}}=-\omega ^{{2}}\cos \omega ti-\omega ^{{2}}\sin \omega tj=-\omega ^{{2}}r\,

Showing that the acceleration of the particle a is opposite to that of r and so it is directed towards the origin.

Again,the magnitude of acceleration =modulus of -\omega ^{{2}}r\, is

\omega ^{{2}}r\, which is proportional to r. i.e the distance of the particle from the origin.

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