Position vector with respect to centre- velocity vector and acceleration vector of a particle in circular motion are r- 3-4- m- v- 4-a- ms-1 and a- -6-b- ms-2 respectively- Speed of particle is constant- Match the following two columns-

We know, for circular motion: r⃗⊥v⃗ So, r⃗.v⃗=0 , #160;since θ =90^0 ⇒ 3 × 4 + ( 4) × ( a) = 0 #160; ⇒ a = 3 And, at the same time, ...

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Standard XII

Physics

Angular Displacement

Position vect...

Question

Position vector (with respect to centre), velocity vector and acceleration vector of a particle in circular motion are $\to r = (3^i - 4^j) m, \to v = (4^i - a^j) m s^{- 1} a n d \to a = (- 6^i + b^j) m s^{- 2}$ respectively. Speed of particle is constant. Match the following two columns.

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Solution

We know, for circular motion: $\to r ⊥ \to v$
So, $\to r . \to v = 0$ , since $θ = 90^{0}$
$\Rightarrow 3 \times 4 + (- 4) \times (- a) = 0$
$\Rightarrow a = - 3$
And, at the same time, $\to v . \to a = 0$ , since $θ = 90^{0}$
$\Rightarrow 4 \times (- 6) + (- a) \times (- b) = 0$
$\Rightarrow - 24 - 3 b = 0$
$\Rightarrow b = - 8$
Radius vector $| \to r | = \sqrt{3^{2} + (- 4)^{2}} = 5$
$\to r . (\to v \times \to a)$
Since $(\to v \times \to a) ⊥ \to r$ , we get $\to r . (\to v \times \to a) = 0$

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Position vector (with respect to centre), velocity vector and acceleration vector of a particle in circular motion are →r=(3^i−4^j)m,→v=(4^i−a^j)ms−1and→a=(−6^i+b^j)ms−2 respectively. Speed of particle is constant. Match the following two columns.

Position vector (with respect to centre), velocity vector and acceleration vector of a particle in circular motion are $\to r = (3^i - 4^j) m, \to v = (4^i - a^j) m s^{- 1} a n d \to a = (- 6^i + b^j) m s^{- 2}$ respectively. Speed of particle is constant. Match the following two columns.