Q. A particle moves in the $x - y$ plane with velocity $v_{x} = 8t - 2$ and $v_{y} = 2$. If it passes through the pointy $x = 14$ and $y = 4$ at $t = 2s$, find the equation ($x - y$ relation) of the path
A) $x = y^{2} - y + 2$
B) $x = 2y^{2} + 2y - 3$
C) $x = 3y^{2} + 5$
D) Can not be found from above data
Explanation
(a) $v_{x} = 8t - 2$
or $\frac{dx}{dt} = 8t - 2$
or $\int_{14}^{x}{dx = \int_{2}^{t}{(8t - 2)dt}}$
or $x - 14 = \left\lbrack 4t^{2} - 2t \right\rbrack_{2}^{t} = 4t^{2} - 2t - 12$
or $x = 4t^{2} - 2t + 2$ …(i)
Further, $v_{y} = 2$
or $\frac{dy}{dt} = 2$
$\therefore\ \ \int_{4}^{y}{dy = \int_{2}^{t}{2dt}}$
or $y - 4 = \lbrack 2t\rbrack_{2}^{t} = 2t - 4$
or $y = 2t$
or $t = \frac{y}{2}$ …(ii)
Substituting value of $t$ from Eq. (ii) in Eq. (i), we have
$x = y^{2} - y + 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.