Question

A particle is moving in X-Y plane such that $v_x=4+4t$ and $v_y=4t.$ If the initial position of the particle is (1, 2). Then the equation of trajectory will be

• A. $y^2=4x$
• B. $y=2x$
• C. $x^2=\frac{y}{2}$
• D. None of these

Answer 1

The correct option is D None of these

$\frac{dx}{dt}=4+4t$. Integrate to get, $x=4t+2t^2+C_1$
x=1 at t=0. Thus, $C_1=1$
$\frac{dy}{dt}=4t$. Integrate to get, $y=2t^2+C_2$
y=2 at t=0. Thus, $C_2=2$
Thus, $y=2t^2+2$ or $t=\sqrt{\frac{y}{2}-1}$
Put in x, $x=4\sqrt{\frac{y}{2}-1}+2(\frac{y}{2}-1)+1$, which is none of the given options.