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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