A point moves in the x - y plane according to x = Asin and y = A (1 – ) where A and ω are positive constants. The distance traversed by the point during the time τ& is

Correct answer: C. Aωτ Distance travelled by point in time t is: S = ∫ _ 0 ^ τ | → V |dt ∵ | → V | = V _ x ^ 2 + V _ y ^ 2 Where, V _ x = (dx)/(dt) = Aω cos and V _ y = (dy)/(dt) = Aω sin S = ∫ _ 0 ^ τ A = Aωτ

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Motion in One Dimension

NTA Abhyas 2020 · Physics · Motion in One Dimension

A point moves in the x - y plane according to x = Asin and y = A (1...

A point moves in the

x - y

plane according to

x = Asin\omegat

and

y = A

\left(1 – \text\cos\omegat\right)

where

and \omega are positive constants. The distance traversed by the point during the time

τ& is

Options

A $\dfrac{A{\omega}^{2}}{τ}$
B $A{\omega}^{2}τ$
C $A\omegaτ$
D $\dfrac{A\omega}{τ}$

Correct answer

C. $A\omegaτ$

Step-by-step solution

Distance travelled by point in time t is:

S = {∫}_{0}^{τ}\left|\overset{→}{V}\right|dt

∵ \left|\overset{→}{V}\right| = \sqrt{{{V}_{x}}^{2} + {{V}_{y}}^{2}}

Where,

{V}_{x} = \dfrac{dx}{dt} = A\omega cos \omegat and{V}_{y} = \dfrac{dy}{dt} = A\omega sin \omegat

S = {∫}_{0}^{τ}A\omegadt = A\omegaτ

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