A particle moves so that its position vector varies with time as r = A cos ω t i ^ + A sin ω t j ^ . The initial velocity of the particle will be :

r = A cos ω t i ^ + A sin ω t j ^ V = d t d r ^ = − ω A sin ω t i ^ + A ω cos ω t j ^ Initial velocity ( t = 0 ) ⇒ V = − ω A sin 0 ∘ i ^ + A ω cos 0 ∘ j ^ v = A ω j ^

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PhysicsKinematics2-D & 3DEasy2 minQB

PhysicsEasysingle choice

Passage / Comprehension

Let the particle 'P' move along a curve between two positions, 1 and 2, at times $t = t_{1}$ and $t = t_{2}$ , respectively. During the time interval $Δ t = (t_{2} - t_{1})$ , the position vector of the particle changes from $r_{1}$ to $r_{2}$ by $Δ r$ . Hence, on average, the position vector changes with time at a rate:

$v_{a v} = \frac{Δ r}{Δ t} = \frac{r _{2} - r _{1}}{t _{2} - t _{1}}$

For the instantaneous velocity, as $Δ t \to 0$ :

$v = Δ t \to 0 lim v_{a v} = Δ t \to 0 lim \frac{Δ r}{Δ t} = \frac{d r}{d t}$

A particle moves so that its position vector varies with time as $r = A cos ω t \hat{i} + A sin ω t \hat{j}$ . The initial velocity of the particle will be :

Options:

Answer:

Solution:

$r = A cos ω t \hat{i} + A sin ω t \hat{j}$

$V = \frac{d r ^}{d t} = - ω A sin ω t \hat{i} + A ω cos ω t \hat{j}$

Initial velocity $(t = 0) \Rightarrow V = - ω A sin 0^{\circ} \hat{i} + A ω cos 0^{\circ} \hat{j}$

$v = A ω \hat{j}$

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:2-D & 3D

⏱ 2mℹ️ Source: QB

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