A particle is moving with a uniform speed v in a circular path of radius r with the centre at O. When the particle moves from a point P to Q on the circle such that angle POQ=θ, then the magnitude of the change in velocity is

Question

A particle is moving with a uniform speed v in a circular path of radius r with the centre at O. When the particle moves from a point P to Q on the circle such that angle POQ=θ, then the magnitude of the change in velocity is (A) 2v sin(2θ) (B) zero (C) 2v sin((θ/2)) (D) 2v cos((θ/2))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/37603b4aa10fe40d08eb9f55e17a511e-.png" /> ∴ Change in magnitude of velocity |Δ v | =√v2+v2-2 v2 cos θ =2 v sin (θ/2)

Q. A particle is moving with a uniform speed $v$ in a circular path of radius $r$ with the centre at $O$ . When the particle moves from a point $P$ to $Q$ on the circle such that $∠ POQ = θ$ , then the magnitude of the change in velocity is

$2 v sin (2 θ)$

zero

$2 v sin (\frac{θ}{2})$

$2 v cos (\frac{θ}{2})$

$∴$ Change in magnitude of velocity
$∣Δ v ∣ = v^{2} + v^{2} - 2 v^{2} cos θ$
$= 2 v sin \frac{θ}{2}$

Q. A particle is moving with a uniform speed v in a circular path of radius r with the centre at O. When the particle moves from a point P to Q on the circle such that ∠POQ=θ, then the magnitude of the change in velocity is

2vsin(2θ)

zero

2vsin(2θ​)

2vcos(2θ​)

∴ Change in magnitude of velocity ∣Δv∣=v2+v2−2v2cosθ​ =2vsin2θ​

Q. A particle is moving with a uniform speed $v$ in a circular path of radius $r$ with the centre at $O$ . When the particle moves from a point $P$ to $Q$ on the circle such that $∠ POQ = θ$ , then the magnitude of the change in velocity is

$2 v sin (2 θ)$

$2 v sin (\frac{θ}{2})$

$2 v cos (\frac{θ}{2})$

$∴$ Change in magnitude of velocity
$∣Δ v ∣ = v^{2} + v^{2} - 2 v^{2} cos θ$
$= 2 v sin \frac{θ}{2}$