A particle is moving with velocity \vec{v}=K(y\hat{i}+x\hat{j}),where\; K is a constant. The general equation for its path is

  • Option 1)

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

  • Option 2)

    y=x^{2}+constant

  • Option 3)

    y^{2}=x+constant

  • Option 4)

    xy=constant    

 

Answers (1)

As we learnt in

Uniform velocity,Non uniform Velocity -

Uniform velocity

If equal displacement are covered in equal intervals of time.

Non uniform velocity

If equal displacement are covered in unequal intervals of time.

- wherein

More to know :

For uniform motion along a straight line, the average and instantaneous velocities have the same values.

 

 \vec{V}=K \left ( yi+xj \right ) - \left ( i \right )

\vec{V}=kyi+kxj \Rightarrow \vec{V}= \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} -\left ( ii \right )

equation (i) and (ii)

\frac{dx}{dt}=ky

\frac{dy}{dt}=kx

\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}

\frac{dy}{dx}=\frac{kx}{ky}=\frac{x}{y} -\left ( iii \right )

integrating both sides of above equation

\int ydy= \int xdx\Rightarrow y ^2=x^2+constant

 


Option 1)

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

correct

Option 2)

y=x^{2}+constant

incorrect

Option 3)

y^{2}=x+constant

incorrect

Option 4)

xy=constant    

incorrect

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions