The position vector of a particle is r = (a cos ω t)i + (a sin ω t)j. The velocity of the particle is (a) directed towards the origin (b) directed away from the origin (c) parallel to the position vector (d) perpendicular to the position vector

Answer: (d). v = (d r)/(dt) = -aωsin ω t i + aωcos ω t j v· r = -a²ωsin ω tcos ω t + a²ωsin ω tcos ω t = 0 So the answer is (d): velocity is perpendicular to the position vector.

The position vector of a particle is r = (a cos ω t)i + (a sin ω t)j… | NEET 1995 (AIPMT)

Q27NEET 1995 · AIPMTEasy

Motion in a Plane (2-D Kinematics)

The position vector of a particle is $\vec{r} = (a \cos \omega t)\hat{i} + (a \sin \omega t)\hat{j}$ . The velocity of the particle is

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Answer: (d)

$\vec v = \frac{d\vec r}{dt} = -a\omega\sin\omega t\,\hat i + a\omega\cos\omega t\,\hat j$
$\vec v\cdot\vec r = -a^2\omega\sin\omega t\cos\omega t + a^2\omega\sin\omega t\cos\omega t = 0$
So the answer is (d): velocity is perpendicular to the position vector.

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Motion in a Plane (2-D Kinematics) question from NEET 1995 (AIPMT)

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