Question

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

Answer 1

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.