The position vector of a particle is r - a -cos- t - - sin- t - The velocity of the particle isparallel to position vector nbsp-directed towards origindirected away from originperpendicular to position vector- nbsp-

The correct option is D perpendicular to position vector. nbsp;r = a {(cos ω t)î + (sinω t)ĵ} ∴ nbsp; v = a{ sinω t ×ωî + cosω t ×ωĵ} r. v = a^2 ωcosω t × ...

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Standard XII

Physics

Uniform Circular Motion

The position ...

Question

The position vector of a particle is $\to r = a [cos ω t^i + sin ω t^j]$ . The velocity of the particle is

parallel to position vector

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directed towards origin

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directed away from origin

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perpendicular to position vector.

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Solution

The correct option is D perpendicular to position vector.
$→r=a{(cosωt)^i+(sinωt)^j}$
$∴→v=a{−sinωt×ω^i+cosωt×ω^j}$
$\to r . \to v = a^{2} ω cos ω t \times - sin ω t$
$+ a^{2} ω sin ω t \times cos ω t = 0$
$∴ \to r ⊥ \to v$

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The position vector of a particle is →r=a[cosωt^i+sinωt^j]. The velocity of the particle is

The correct option is D perpendicular to position vector. →r=a{(cosωt)^i+(sinωt)^j} ∴→v=a{−sinωt×ω^i+cosωt×ω^j} →r.→v=a2ωcosωt×−sinωt +a2ωsinωt×cosωt=0 ∴→r⊥→v

The position vector of a particle is $\to r = a [cos ω t^i + sin ω t^j]$ . The velocity of the particle is

The correct option is D perpendicular to position vector.
$→r=a{(cosωt)^i+(sinωt)^j}$
$∴→v=a{−sinωt×ω^i+cosωt×ω^j}$
$\to r . \to v = a^{2} ω cos ω t \times - sin ω t$
$+ a^{2} ω sin ω t \times cos ω t = 0$
$∴ \to r ⊥ \to v$