Physics, asked by anshikabajpayee03, 10 months ago

a particle moves in the x-y plane from origin having instantaneous velocity v`=2i+4xj. the equation of path is​

Answers

Answered by sonuvuce
3

Answer:

y=2x^2

Explanation:

Given the instantaneous velocity of the particle

v=2\hat i+4x\hat j

\implies \frac{dx}{dt}=2

\implies dx=2dt

\implies \int_0^x=2\int_0^t dt

\implies x=2t

\implies t=\frac{x}{2}

And

\frac{dy}{dt}=4x

\implies dy=4xdt

\implies \int_0^y=4x\int_0^t

\implies y=4xt

\implies y=4x\times\frac{x}{2}

\implies y=2x^2

This is the equation of the path.

Hope this helps.

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