Question

A particle moves in the $x-y$ plane with velocity $v_x=8t-2$ and $v_y=2.$ If it passes through the point $x=14$ and $y=4$ at $t=2s,$ the equation of the trajectory is

• A. $x=y^2-y+2$
• B. $x=y^2-2$
• C. $x=y^2+y-6$
• D. None of these

Answer 1

The correct option is A $x=y^2-y+2$

$v_x=8t-2$ or, $\frac{dx}{dt}=8t-2$
Integrate to get, $x=4t^2-2t+c$
Given, at $t=2,x=14,$ thus $14=4\left(4\right)-2\left(2\right)+c$ or $c=2$
Thus, $x=4t^2-2t+2$
Similarly, $v_y=2$ or $\frac{dy}{dt}=2$
Integrate to get, $y=2t+c^{\prime }$
Given, at $t=2,y=4,$ thus $2\left(2\right)+c^{\prime }=4$ or $c^{\prime }=0$
Thus, $y=2t($or $t=\frac{y}{2})$
Put this $t$ in $x$ to get $x=4{\left(\frac{y}{2}\right)}^2-2\left(\frac{y}{2}\right)+2=y^2-y+2$