A particle is moving with a velocity → v = K (y ^ i + x ^ j ), where K is a constant. The general equation for its path is:

Correct answer: D. y ^ 2 = x ^ 2 + constant → v = K(y ^ i + x ^ j ) v _ x = y = (dx)/(dt) v _ y = x = (dy)/(dt) (dx)/(dy) = (y)/(x) ∫ xdx = ∫ ydy x ^ 2 2 = y ^ 2 2 + C or y ^ 2 = x ^ 2 + constant

JEE Main 20199 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Morning Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2019 · 9 January · Morning Shift · Physics · Mathematics in Physics

A particle is moving with a velocity → v = K (y ^ i + x ^ j ),...

A particle is moving with a velocity

\overset{\to}{v} = K

\left(y\overset{^}{i} + x\overset{^}{j}\right),

where

K is a constant.

The general equation for its path is:

Options

A ${y}^{2} = x + constant$
B $xy = constant$
C $y = {x}^{2} + constant$
D ${y}^{2} = {x}^{2} + constant$

Correct answer

D. ${y}^{2} = {x}^{2} + constant$

Step-by-step solution

\overset{\to}{v} = K\left(y\overset{^}{i} + x\overset{^}{j}\right)

{v}_{x} = y = \dfrac{dx}{dt}

{v}_{y} = x = \dfrac{dy}{dt}

\dfrac{dx}{dy} = \dfrac{y}{x}

∫ xdx = ∫ ydy

\dfrac{{x}^{2}}{2} = \dfrac{{y}^{2}}{2} + C

{y}^{2} = {x}^{2} + constant

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