The position vector of a particle is r-a cos - t-a sin - t- The velocity of the particle is

R⃗=(a cos ω t)î+(a sin ω t)ĵ v⃗= d⃗r⃗/dt= a ω sin ω t î + a ω cos ω t ĵ As r⃗. v⃗=0 therefore velocity of the particle is perpendicular to the position vecto ...

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Standard XI

Physics

Addition and Subtraction in Unit Vector Notation

The position ...

Question

The position vector of a particle is $\to r = (a c o s ω t)^i + (a s i n ω t)^j$ . The velocity of the particle is

Parallel to the position vector

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Perpendicular to the position vector

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Directed towards the origin

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Directed away from the origin

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Solution

The correct option is B
Perpendicular to the position vector

$\to r = (a c o s ω t)^i + (a s i n ω t)^j$
$\to v = \frac{\to d r}{d t} = - a ω s i n ω t^i + a ω c o s ω t^j$
As $\to r . \to v = 0$ therefore velocity of the particle is perpendicular to the position vector.

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The position vector of a particle is →r=(a cos ωt)^i+(a sin ωt)^j. The velocity of the particle is

The correct option is B Perpendicular to the position vector →r=(a cos ωt)^i+(a sin ωt)^j →v=→drdt=−aω sin ωt^i+aω cos ωt^j As →r.→v=0 therefore velocity of the particle is perpendicular to the position vector.

The position vector of a particle is $\to r = (a c o s ω t)^i + (a s i n ω t)^j$ . The velocity of the particle is

The correct option is B
Perpendicular to the position vector

$\to r = (a c o s ω t)^i + (a s i n ω t)^j$
$\to v = \frac{\to d r}{d t} = - a ω s i n ω t^i + a ω c o s ω t^j$
As $\to r . \to v = 0$ therefore velocity of the particle is perpendicular to the position vector.