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A particle moves so that its position ve...

A particle moves so that its position vector is given by `vecr=cosomegathatx+sinomegathaty`, where `omega` is a constant. Which of the following is true?

A

Velocity and acceleration both are paralled to `vecr`

B

Velocity is perpendicular to `vecr` and acceleration is directed towards the origin

C

Velocity is perpendicular to `vecr` and acceleration is directed away from the origin

D

Velocity and acceleration both are perpendicular to `vecr`

Text Solution

Verified by Experts

The correct Answer is:
B
`vecr=cosomegathatx+sinomegathaty`
Velocity of the particle,
`vecv=(dvecr)/(dt)=-omega(sinomegat)hatx+omega(cosomegat)haty`
Acceleration of the particle,
`veca=(dvecv)/(dt)=-omega^(2)(cosomegat)hatx-omega^(2)(sinomegat)haty`
= `-omega^(2)(cosomegathatx+sinomegathaty)=-omega^(2)vecr`
`therefore` Acceleration is directed towards origin.
Again, `vecr*vecv=-omegasinomegatcosomegat+omegasinomegatcosomegat=0`
Thus, `vecv` is perpendicular to `vecr`.
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