Question

A particle projected from the origin $(x=y=0)$ moves in xy plane so that its velocity is $v=(2i+4xj)$m/s, when it is at point $(x,y)m$, ($i$ and $j$ are the unit vectors along $x$ and $y$ axis). What is the value of $y$, when $x=3$?

Answer 1

As $v=2\hat{i}+4x\hat{j}$ so $v_x=2$ and $v_y=4x$
$v_x=\frac{dx}{dt}=2$
integrating, ${\int }_0^{x}dx={\int }_0^{t}2dt\Rightarrow x=2t...\left(1\right)$
also, $v_y=\frac{dy}{dt}=4x=4\left(2t\right)=8t$
integrating, ${\int }_0^{y}dy={\int }_0^{t}8tdt$
or $y=4t^2$
using (1), $y=4\left(x^2/4\right)=x^2$
when, $x=3,y=3^2=9$