The position vector of the particle is r t = a cos ⁡ ω t i ^ + a sin ⁡ ω t j ^ , where a and ω are real constants of suitable dimensions. The acceleration is

Given that, r t = a cos ⁡ ω t i ^ + a sin ⁡ ω t j ^ ∵ v → = d r t d t = - a ω sin ⁡ ω t i ^ + a ω cos ⁡ ω t i ^ j a → = d v → d t = - a ω 2 cos ⁡ ω t i ^ - a ω 2 cos ⁡ ω t j ^ a → . v → = a 2 ω 3 sin ⁡ ω t cos ⁡ ω t - a 2 ω 3 sin ⁡ ω t cos ⁡ ω t ⇒ a → . v → = 0 Above result implies that acceleration is perpendicular to velocity.

Physics - Kinematics Question with Solution | TestHub

PhysicsKinematicsRelativeEasy2 minPYQ_2020

PhysicsEasysingle choice

The position vector of the particle is $r (t) = a \cos ω t \hat{i} + a \sin ω t \hat{j}$ , where $a$ and $ω$ are real constants of suitable dimensions. The acceleration is

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Solution:

Given that, $r (t) = a \cos ω t \hat{i} + a \sin ω t \hat{j}$

$∵ \vec{v} = \frac{d r (t)}{d t} = - a ω \sin ω t \hat{i} + a ω \cos ω t \hat{i} j$

$\vec{a} = \frac{d \vec{v}}{d t} = - a ω^{2} \cos ω t \hat{i} - a ω^{2} \cos ω t \hat{j}$

$\vec{a} . \vec{v} = a^{2} ω^{3} \sin ω t \cos ω t - a^{2} ω^{3} \sin ω t \cos ω t$

$\Rightarrow$ $\vec{a} . \vec{v} = 0$

Above result implies that acceleration is perpendicular to velocity.

Subject:PhysicsTopic:KinematicsSubtopic:Relative

⏱ 2mℹ️ Source: PYQ_2020

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